case study class 11 trigonometry

CBSE Case Study Questions for Class 11 Maths Sets Free PDF

Mere Bacchon, you must practice the CBSE Case Study Questions Class 11 Maths Sets in order to fully complete your preparation . They are very very important from exam point of view. These tricky Case Study Based Questions can act as a villain in your heroic exams!

I have made sure the questions (along with the solutions) prepare you fully for the upcoming exams. To download the latest CBSE Case Study Questions , just click ‘ Download PDF ’.

CBSE Case Study Questions for Class 11 Maths Sets PDF

Checkout our case study questions for other chapters.

• Chapter 2 Relations and Functions Case Study Questions
• Chapter 3 Trigonometric Functions Case Study Questions
• Chapter 4 Principle of Mathematical Induction Case Study Questions
• Chapter 5 Complex Numbers and Quadratic Equations Case Study Questions

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Practice MCQ Questions (Very Important)

Practice Assertion Reason & Case Study Based Questions

Sit through FULLY INVIGILATED TESTS involving MCQs. Assertion reason & Case Study Based Questions

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Class 11 Mathematics Case Study Questions

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If you’re seeking a comprehensive and dependable study resource with Class 11 mathematics case study questions for CBSE, myCBSEguide is the place to be. It has a wide range of study notes, case study questions, previous year question papers, and practice questions to help you ace your examinations. Furthermore, it is routinely updated to bring you up to speed with the newest CBSE syllabus. So, why delay? Begin your path to success with myCBSEguide now!

The rationale behind teaching Mathematics

The general rationale to teach Mathematics at the senior secondary level is to assist students:

• In knowledge acquisition and cognitive understanding of basic ideas, words, principles, symbols, and mastery of underlying processes and abilities, notably through motivation and visualization.
• To experience the flow of arguments while demonstrating a point or addressing an issue.
• To use the information and skills gained to address issues using several methods wherever possible.
• To cultivate a good mentality in order to think, evaluate, and explain coherently.
• To spark interest in the subject by taking part in relevant tournaments.
• To familiarise pupils with many areas of mathematics utilized in daily life.
• To pique students’ interest in studying mathematics as a discipline.

Case studies in Class 11 Mathematics

A case study in mathematics is a comprehensive examination of a specific mathematical topic or scenario. Case studies are frequently used to investigate the link between theory and practise, as well as the connections between different fields of mathematics. A case study will frequently focus on a specific topic or circumstance and will investigate it using a range of methodologies. These approaches may incorporate algebraic, geometric, and/or statistical analysis.

Sample Class 11 Mathematics case study questions

When it comes to preparing for Class 11 Mathematics, one of the best things Class 11 Mathematics students can do is to look at some Class 11 Mathematics sample case study questions. Class 11 Mathematics sample case study questions will give you a good idea of the types of Class 11 Mathematics sample case study questions that will be asked in the exam and help you to prepare more effectively.

Looking at sample questions is also a good way to identify any areas of weakness in your knowledge. If you find that you struggle with a particular topic, you can then focus your revision on that area.

myCBSEguide offers ample Class 11 Mathematics case study questions, so there is no excuse. With a little bit of preparation, Class 11 Mathematics students can boost their chances of getting the grade they deserve.

Some samples of Class 11 Mathematics case study questions are as follows:

Class 11 Mathematics case study question 1

• 9 km and 13 km
• 9.8 km and 13.8 km
• 9.5 km and 13.5 km
• 10 km and 14 km
• x  ≤   −1913
• x <  −1613
• −1613  < x <  −1913
• There are no solution.
• y  ≤   12 x+2
• y >  12 x+2
• y  ≥   12 x+2
• y <  12 x+2

Answer Key:

• (b) 9.8 km and 13.8 km
• (a) −1913   ≤  x 
• (b)  y >  12 x+2
• (d) (-5, 5)

Class 11 Mathematics case study question 2

• 2 C 1 × 13 C 10
• 2 C 1 × 10 C 13
• 1 C 2 × 13 C 10
• 2 C 10 × 13 C 10
• 6 C 2​ × 3 C 4   × 11 C 5 ​
• 6 C 2​ × 3 C 4   × 11 C 5
• 6 C 2​ × 3 C 5 × 11 C 4 ​
• 6 C 2 ​  ×   3 C 1 ​  × 11 C 5 ​
• (b) (13) 4  ways
• (c) 2860 ways.

Class 11 Mathematics case study question 3

Read the Case study given below and attempt any 4 sub parts: Father of Ashok is a builder, He planned a 12 story building in Gurgaon sector 5. For this, he bought a plot of 500 square yards at the rate of Rs 1000 /yard². The builder planned ground floor of 5 m height, first floor of 4.75 m and so on each floor is 0.25 m less than its previous floor.

Class 11 Mathematics case study question 4

Read the Case study given below and attempt any 4 sub parts: villages of Shanu and Arun’s are 50km apart and are situated on Delhi Agra highway as shown in the following picture. Another highway YY’ crosses Agra Delhi highway at O(0,0). A small local road PQ crosses both the highways at pints A and B such that OA=10 km and OB =12 km. Also, the villages of Barun and Jeetu are on the smaller high way YY’. Barun’s village B is 12km from O and that of Jeetu is 15 km from O.

Now answer the following questions:

• 5x + 6y = 60
• 6x + 5y = 60
• (a) (10, 0)
• (b) 6x + 5y = 60
• (b) 60/√ 61 km
• (d) 2√61 km

A peek at the Class 11 Mathematics curriculum

The Mathematics Syllabus has evolved over time in response to the subject’s expansion and developing societal requirements. The Senior Secondary stage serves as a springboard for students to pursue higher academic education in Mathematics or professional subjects such as Engineering, Physical and Biological Science, Commerce, or Computer Applications. The current updated curriculum has been prepared in compliance with the National Curriculum Framework 2005 and the instructions provided by the Focus Group on Teaching Mathematics 2005 in order to satisfy the rising demands of all student groups. Greater focus has been placed on the application of various principles by motivating the themes from real-life events and other subject areas.

Class 11 Mathematics (Code No. 041)

Design of Class 11 Mathematics exam paper

CBSE Class 11 mathematics question paper is designed to assess students’ understanding of the subject’s essential concepts. Class 11 mathematics question paper will assess their problem-solving and analytical abilities. Before beginning their test preparations, students in Class 11 maths should properly review the question paper format. This will assist Class 11 mathematics students in better understanding the paper and achieving optimum scores. Refer to the Class 11 Mathematics question paper design provided.

 Class 11 Mathematics Question Paper Design

• No chapter-wise weightage. Care to be taken to cover all the chapters.
• Suitable internal variations may be made for generating various templates keeping the overall weightage to different forms of questions and typology of questions the same.  

Choice(s): There will be no overall choice in the question paper. However, 33% of internal choices will be given in all the sections.

  Prescribed Books:

• Mathematics Textbook for Class XI, NCERT Publications
• Mathematics Exemplar Problem for Class XI, Published by NCERT
• Mathematics Lab Manual class XI, published by NCERT

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Chapter 3 Class 11 Trigonometric Functions

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NCERT Solutions of Chapter 3 Class 11 Trigonometry is available free at teachoo. You can check the detailed explanation of all questions of exercises, examples and miscellaneous by clicking on the Exercise link below.

We had learned Basics of Trigonometry in Class 10. In this chapter, we will learn

• What is a positive or a negative angle
• Measuring angles in Degree , Minutes and Seconds
• Radian measure of an angle
• Converting Degree to Radians , and vice-versa
• Sign of sin, cos, tan in all 4 quadrants
• Finding values of trigonometric functions when one value is given (Example: Finding value of sin, cot, cosec, tan, sec, when cos x = -3/5 is given)
• Finding Value of trigonometric functions, given angle
• Solving questions by formula like  (x + y) formula, 2x 3x formula, Cos x + cos y formula , 2 sin x sin y formula 
• Finding principal and general solutions of a trigonometric equation
• Sin and Cosine Formula with supplementary Questions

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CBSE Class 11 Maths – Chapter 3 Trigonometric Functions- Study Materials

NCERT Solutions Class 11 All Subjects Sample Papers Past Years Papers

Sets : Notes and Study Materials -pdf

• Concepts of  Trigonometric Functions
• Trigonometric Functions Master File
• Trigonometric Functions Revision Notes
• R D Sharma Solution of Trigonometric Functions
• NCERT Solution  Trigonometric Functions
• NCERT  Exemplar Solution Trigonometric Functions
• Trigonometric Functions : Solved Example 1

CBSE Class 11 Maths Notes Chapter 3 Trigonometric Functions

Angle Angle is a measure of rotation of a given ray about its initial point. The original ray is called the initial side and the final position of ray after rotation is called terminal side of the angle. The point of rotation is called vertex. If the direction of rotation is anti-clockwise, the angle is said to be positive and if the direction of rotation is clockwise, then the angle is negative.

Measuring Angles There are two systems of measuring angles Sexagesimal system (degree measure): If a rotation from the initial side to terminal side is  ( 1 360 ) t h  of a revolution, the angle is said to have a measure of one degree, written as 1°. One sixtieth of a degree is called a minute, written as 1′ and one-sixtieth of a minute is called a second, written as 1″ Thus, 1° = 60′ and 1′ = 60″

Circular system (radian measure):  A radian is an angle subtended at the centre of a circle by an arc, whose length is equal to the radius of the circle. We denote 1 radian by 1°.

Relation Between Radian and Degree We know that a complete circle subtends at its centre an angle whose measure is 2π radians as well as 360°. 2π radian = 360°. Hence, π radian = 180° or 1 radian = 57° 16′ 21″ (approx) 1 degree = 0.01746 radian

Six Fundamental Trigonometric Identities

• sinx =  1 c o s e c x
• cos x =  1 s e c x
• tan x =  1 c o t x
• sin 2  x + cos 2  x = 1
• 1 + tan 2 x = sec 2  x
• 1 + cot 2  x = cosec 2  x

Trigonometric Functions – Class 11 Maths Notes

Trigonometric ratios are defined for acute angles as the ratio of the sides of a right angled triangle. The extension of trigonometric ratios to any angle in terms of radian measure (real number) are called trigonometric function. The signs of trigonometric function in different quadrants have been given in following table.

Domain and Range of Trigonometric Functions

Sine, Cosine, and Tangent of Some Angles Less Than 90°

Allied or Related Angles The angles  n π 2 ± θ  are called allied or related angle and θ ± n × (2π) are called coterminal angles. For general reduction, we have following rules, the value of trigonometric function for ( n π 2 ± θ ) is numerically equal to

• the value of the same function, if n is an even integer with the algebraic sign of the function as per the quadrant in which angle lies.
• the corresponding co-function of θ, if n is an odd integer with the algebraic sign of the function for the quadrant in which it lies, here sine and cosine, tan and cot, sec and cosec are cofunctions of each other.

Functions of Negative Angles

For any acute angle of θ. We have,

• sin(-θ) = – sinθ
• cos (-θ) = cosθ
• tan (-θ) = – tanθ
• cot (-θ) = – cotθ
• sec (-θ) = secθ
• cosec (-θ) = – cosecθ

Some Formulae Regarding Compound Angles

An angle made up of the sum or difference of two or more angles is called compound angles. The basic results in direction are called trigonometric identities as given below: (i) sin (x + y) = sin x cos y + cos x sin y (ii) sin (x – y) = sin x cos y – cos x sin y (iii) cos (x + y) = cos x cos y – sin x sin y (iv) cos (x – y) = cos x cos y + sin x sin y

(ix) sin(x + y) sin (x – y) = sin 2  x – sin 2  y = cos 2  y – cos 2  x (x) cos (x + y) cos (x – y) = cos 2  x – sin 2  y = cos 2  y – sin 2  x

Transformation Formulae

• 2 sin x cos y = sin (x + y) + sin (x – y)
• 2 cos x sin y = sin (x + y) – sin (x – y)
• 2 cos x cos y = cos (x + y) + cos (x – y)
• 2 sin x sin y = cos (x – y) – cos (x + y)
• sin x + sin y = 2 sin( x + y 2 ) cos( x − y 2 )
• sin x – sin y = 2 cos( x + y 2 ) sin( x − y 2 )
• cos x + cos y = 2 cos( x + y 2 ) cos( x − y 2 )
• cos x – cos y = -2 sin( x + y 2 ) sin( x − y 2 )

Trigonometric Ratios of Multiple Angles

Product of Trigonometric Ratios

• sin x sin (60° – x) sin (60° + x) =  1 4  sin 3x
• cos x cos (60° – x) cos (60° + x) =  1 4  cos 3x
• tan x tan (60° – x) tan (60° + x) = tan 3x
• cos 36° cos 72° =  1 4
• cos x . cos 2x . cos 2 2 x . cos 2 3 x … cos 2 n-1  =  s i n 2 n x 2 n s i n x

Sum of Trigonometric Ratio, if Angles are in A.P.

Trigonometric Equations Equation which involves trigonometric functions of unknown angles is known as the trigonometric equation.

Solution of a Trigonometric Equation A solution of a trigonometric equation is the value of the unknown angle that satisfies the equation. A trigonometric equation may have an infinite number of solutions.

Principal Solution The solutions of a trigonometric equation for which 0 ≤ x ≤ 2π are called principal solutions.

General Solutions A solution of a trigonometric equation, involving ‘n’ which gives all solution of a trigonometric equation is called the general solutions.

General Solutions of Trigonometric Equation

• sin x = 0 ⇔ x = nπ, n ∈ Z
• cos x = 0 ⇔ x = (2n + 1)  π 2  , n ∈ Z
• tan x = 0 ⇔ x = nπ, n ∈ Z
• sin x = sin y ⇔ x = nπ + (-1) n  y, n ∈ Z
• cos x = cos y ⇔ x = 2nπ ± y, n ∈ Z
• tan x = tan y ⇔ x = nπ ± y, n ∈ Z
• sin 2  x = sin 2  y ⇔ x = nπ ± y, n ∈ Z
• cos 2  x = cos 2  y ⇔ x = nπ ± y, n ∈ Z
• tan 2  x = tan 2  y ⇔ x = nπ ± y, n ∈ Z

Basic Rules of Triangle

In a triangle ABC, the angles are denoted by capital letters A, B and C and the lengths of sides of opposite to these angles are denoted by small letters a, b and c, respectively. Sine Rule s i n A a = s i n B b = s i n C c

Cosine Rule a 2  = b 2  + c 2  – 2bc cos A b 2  = c 2  + a 2  – 2ac cos B c 2  = a 2  + b 2  – 2ab cos C

Projection Rule a = b cos C + c cos B b = c cos A + a cos C c = a cos B + b cos A

Trigonometric Functions Class 11 MCQs Questions with Answers

Question 1. The value of cos² x + cos² y – 2cos x × cos y × cos (x + y) is (a) sin (x + y) (b) sin² (x + y) (c) sin³ (x + y) (d) sin 4 (x + y)

Answer: (b) sin² (x + y) Hint: cos² x + cos² y – 2cos x × cos y × cos(x + y) {since cos(x + y) = cos x × cos y – sin x × sin y } = cos² x + cos² y – 2cos x × cos y × (cos x × cos y – sin x × sin y) = cos² x + cos² y – 2cos² x × cos² y + 2cos x × cos y × sin x × sin y = cos² x + cos² y – cos² x × cos² y – cos² x × cos² y + 2cos x × cos y × sin x × sin y = (cos² x – cos² x × cos² y) + (cos² y – cos² x × cos² y) + 2cos x × cos y × sin x × sin y = cos² x(1- cos² y) + cos² y(1 – cos² x) + 2cos x × cos y × sin x × sin y = sin² y × cos² x + sin² x × cos² y + 2cos x × cos y × sin x × sin y (since sin² x + cos² x = 1 ) = sin² x × cos² y + sin² y × cos² x + 2cos x × cos y × sin x × sin y = (sin x × cos y)² + (sin y × cos x)² + 2cos x × cos y × sin x × sin y = (sin x × cos y + sin y × cos x)² = {sin (x + y)}² = sin² (x + y)

Question 2. If a×cos x + b × cos x = c, then the value of (a × sin x – b²cos x)² is (a) a² + b² + c² (b) a² – b² – c² (c) a² – b² + c² (d) a² + b² – c²

Answer: (d) a² + b² – c² Hint: We have (a×cos x + b × sin x)² + (a × sin x – b × cos x)² = a² + b² ⇒ c² + (a × sin x – b × cos x)² = a² + b² ⇒ (a × sin x – b × cos x)² = a² + b² – c²

Question 3. If cos a + 2cos b + cos c = 2 then a, b, c are in (a) 2b = a + c (b) b² = a × c (c) a = b = c (d) None of these

Answer: (a) 2b = a + c Hint: Given, cos A + 2 cos B + cos C = 2 ⇒ cos A + cos C = 2(1 – cos B) ⇒ 2 cos((A + C)/2) × cos((A-C)/2 = 4 sin²(B/2) ⇒ 2 sin(B/2)cos((A-C)/2) = 4sin² (B/2) ⇒ cos((A-C)/2) = 2sin (B/2) ⇒ cos((A-C)/2) = 2cos((A+C)/2) ⇒ cos((A-C)/2) – cos((A+C)/2) = cos((A+C)/2) ⇒ 2sin(A/2)sin(C/2) = sin(B/2) ⇒ 2{√(s-b)(s-c)√bc} × {√(s-a)(s-b)√ab} = √(s-a)(s-c)√ac ⇒ 2(s – b) = b ⇒ a + b + c – 2b = b ⇒ a + c – b = b ⇒ a + c = 2b

Question 4. The value of cos 5π is (a) 0 (b) 1 (c) -1 (d) None of these

Answer: (c) -1 Hint: Given, cos 5π = cos (π + 4π) = cos π = -1

Question 5. In a triangle ABC, cosec A (sin B cos C + cos B sin C) equals (a) none of these (b) c/a (c) 1 (d) a/c

Answer: (c) 1 Hint: Given cosec A (sin B cos C + cos B sin C) = cosec A × sin(B+C) = cosec A × sin(180 – A) = cosec A × sin A = cosec A × 1/cosec A = 1

Question 6. If the angles of a triangle be in the ratio 1 : 4 : 5, then the ratio of the greatest side to the smallest side is (a) 4 : (√5 – 1) (b) 5 : 4 (c) (√5 – 1) : 4 (d) none of these

Answer: (a) 4 : (√5 – 1) Hint: Given, the angles of a triangle be in the ratio 1 : 4 : 5 ⇒ x + 4x + 5x = 180 ⇒ 10x = 180 ⇒ x = 180/10 ⇒ x = 18 So, the angle are: 18, 72, 90 Since a : b : c = sin A : sin B : sin C ⇒ a : b : c = sin 18 : sin 72 : sin 90 ⇒ a : b : c = (√5 – 1)/4 : {√(10 + 2√5)}/4 : 1 ⇒ a : b : c = (√5 – 1) : {√(10 + 2√5)} : 4 Now, c /a = 4/(√5 – 1) ⇒ c : a = 4 : (√5 – 1)

Question 7. The value of cos 180° is (a) 0 (b) 1 (c) -1 (d) infinite

Answer: (c) -1 Hint: 180 is a standard degree generally we all know their values but if we want to go theoretically then cos(90 + x) = – sin(x) So, cos 180 = cos(90 + 90) = -sin 90 = -1 {sin 90 = 1} So, cos 180 = -1

Question 8. The perimeter of a triangle ABC is 6 times the arithmetic mean of the sines of its angles. If the side b is 2, then the angle B is (a) 30° (b) 90° (c) 60° (d) 120°

Answer: (b) 90° Hint: Let the lengths of the sides if ∆ABC be a, b and c Perimeter of the triangle = 2s = a + b + c = 6(sinA + sinB + sinC)/3 ⇒ (sinA + sinB + sinC) = ( a + b + c)/2 ⇒ (sinA + sinB + sinC)/( a + b + c) = 1/2 From sin formula,Using sinA/a = sinB/b = sinC/c = (sinA + sinB + sinC)/(a + b + c) = 1/2 Now, sinB/b = 1/2 Given b = 2 So, sinB/2 = 1/2 ⇒ sinB = 1 ⇒ B = π/2

Question 9: If 3 × tan(x – 15) = tan(x + 15), then the value of x is (a) 30 (b) 45 (c) 60 (d) 90

Answer: (b) 45 Hint: Given, 3×tan(x – 15) = tan(x + 15) ⇒ tan(x + 15)/tan(x – 15) = 3/1 ⇒ {tan(x + 15) + tan(x – 15)}/{tan(x + 15) – tan(x – 15)} = (3 + 1)/(3 – 1) ⇒ {tan(x + 15) + tan(x – 15)}/{tan(x + 15) – tan(x – 15)} = 4/2 ⇒ {tan(x + 15) + tan(x – 15)}/{tan(x + 15) – tan(x – 15)} = 2 ⇒ sin(x + 15 + x – 15)/sin(x + 15 – x + 15) = 2 ⇒ sin 2x/sin 30 = 2 ⇒ sin 2x/(1/2) = 2 ⇒ 2 × sin 2x = 2 ⇒ sin 2x = 1 ⇒ sin 2x = sin 90 ⇒ 2x = 90 ⇒ x = 45

Question 10. If the sides of a triangle are 13, 7, 8 the greatest angle of the triangle is (a) π/3 (b) π/2 (c) 2π/3 (d) 3π/2

Answer: (c) 2π/3 Hint: Given, the sides of a triangle are 13, 7, 8 Since greatest side has greatest angle, Now Cos A = (b² + c² – a²)/2bc ⇒ Cos A = (7² + 8² – 13²)/(2×7×8) ⇒ Cos A = (49 + 64 – 169)/(2×7×8) ⇒ Cos A = (113 – 169)/(2×7×8) ⇒ Cos A = -56/(2×56) ⇒ Cos A = -1/2 ⇒ Cos A = Cos 2π/3 ⇒ A = 2π/3 So, the greatest angle is = 2π/3

Question 11. The value of tan 20 × tan 40 × tan 80 is (a) tan 30 (b) tan 60 (c) 2 tan 30 (d) 2 tan 60

Answer: (b) tan 60 Hint: Given, tan 20 × tan 40 × tan 80 = tan 40 × tan 80 × tan 20 = [{sin 40 × sin 80}/{cos 40 × cos 80}] × (sin 20/cos 20) = [{2 * sin 40 × sin 80}/{2 × cos 40 × cos 80}] × (sin 20/cos 20) = [{cos 40 – cos 120}/{cos 120 + cos 40}] × (sin 20/cos 20) = [{cos 40 – cos (90 + 30)}/{cos (90 + 30) + cos 40}] × (sin 20/cos 20) = [{cos 40 + sin30}/{-sin30 + cos 40}] × (sin 20/cos 20) = [{(2 × cos 40 + 1)/2}/{(-1 + cos 40)/2}] × (sin 20/cos 20) = [{2 × cos 40 + 1}/{-1 + cos 40}] × (sin 20/cos 20) = [{2 × cos 40 × sin 20 + sin 20}/{-cos 20 + cos 40 × cos 20}] = (sin 60 – sin 20 + sin 20)/(-cos 20 + cos 60 + cos 20) = sin 60/cos 60 = tan 60 So, tan 20 × tan 40 × tan 80 = tan 60

Question 12. If the angles of a triangle be in the ratio 1 : 4 : 5, then the ratio of the greatest side to the smallest side is (a) 4 : (√5 – 1) (b) 5 : 4 (c) (√5 – 1) : 4 (d) none of these

Question 13. The general solution of √3 cos x – sin x = 1 is (a) x = n × π + (-1)n × (π/6) (b) x = π/3 – n × π + (-1)n × (π/6) (c) x = π/3 + n × π + (-1)n × (π/6) (d) x = π/3 – n × π + (π/6)

Answer: (c) x = π/3 + n × π + (-1)n × (π/6) Hint: √3 cos x-sin x=1 ⇒ (√3/2)cos x – (1/2)sin x = 1/2 ⇒ sin 60 × cos x – cos 60 × sin x = 1/2 ⇒ sin (x – 60) = 1/2 ⇒ sin (x – π/3) = sin 30 ⇒ sin (x – π/3) = sinπ/6 ⇒ x – π/3 = n × π + (-1)n × (π/6) {where n ∈ Z} ⇒ x = π/3 + n × π + (-1)n × (π/6)

Question 14. If tan² θ = 1 – e², then the value of sec θ + tan³ θ × cosec θ is (a) 2 – e² (b) (2 – e²) 1/2 (c) (2 – e²)² (d) (2 – e²) 3/2

Question 15. The value of cos 20 + 2sin² 55 – √2 sin65 is (a) 0 (b) 1 (c) -1 (d) None of these

Answer: (b) 1 Hint: Given, cos 20 + 2sin² 55 – √2 sin65 = cos 20 + 1 – cos 110 – √2 sin65 {since cos 2x = 1 – 2sin² x} = 1 + cos 20 – cos 110 – √2 sin65 = 1 – 2 × sin {(20 + 110)/2 × sin{(20 – 110)/2} – √2 sin65 {Apply cos C – cos D formula} = 1 – 2 × sin 65 × sin (-45) – √2 sin65 = 1 + 2 × sin 65 × sin 45 – √2 sin65 = 1 + (2 × sin 65)/√2 – √2 sin65 = 1 + √2 ( sin 65 – √2 sin 65 = 1 So, cos 20 + 2sin² 55 – √2 sin65 = 1

Question 16. If the radius of the circumcircle of an isosceles triangle PQR is equal to PQ ( = PR), then the angle P is (a) 2π/3 (b) π/3 (c) π/2 (d) π/6

Answer: (a) 2π/3 Hint: Let S be the center of the circumcircle of triangle PQR. So, SP = SQ = SR = PQ = PR, where SP, SQ & SR are radii. Thus SPQ & SPR are equilateral triangles. ⇒ ∠QSP = 60°; Similarly ∠RQP = 60° ⇒ Angle at the center QSP = 120° So, SRPQ is a rhombus, since all the four sides are equal. Hence, its opposite angles are equal; so ∠P = ∠QSP = 120°

Question 17. If cos a + 2cos b + cos c = 2 then a, b, c are in (a) 2b = a + c (b) b² = a × c (c) a = b = c (d) None of these

Answer: (a) 2b = a + c Hint: Given, cos A + 2 cos B + cos C = 2 ⇒ cos A + cos C = 2(1 – cos B) ⇒ 2 cos((A + C)/2) × cos((A-C)/2 = 4 sin² (B/2) ⇒ 2 sin(B/2)cos((A-C)/2) = 4sin² (B/2) ⇒ cos((A-C)/2) = 2sin (B/2) ⇒ cos((A-C)/2) = 2cos((A+C)/2) ⇒ cos((A-C)/2) – cos((A+C)/2) = cos((A+C)/2) ⇒ 2sin(A/2)sin(C/2) = sin(B/2) ⇒ 2{√(s-b)(s-c)√bc} × {√(s-a)(s-b)√ab} = √(s-a)(s-c)√ac ⇒ 2(s – b) = b ⇒ a + b + c – 2b = b ⇒ a + c – b = b ⇒ a + c = 2b

Question 18. The value of 4 × sin x × sin(x + π/3) × sin(x + 2π/3) is (a) sin x (b) sin 2x (c) sin 3x (d) sin 4x

Answer: (c) sin 3x Hint: Given, 4 × sin x × sin(x + π/3) × sin(x + 2π/3) = 4 × sin x × {sin x × cos π/3 + cos x × sin π/3} × {sin x × cos 2π/3 + cos x × sin 2π/3} = 4 × sin x × {(sin x)/2 + (√3 × cos x)/2} × {-(sin x)/2 + (√3 × cos x)/2} = 4 × sin x × {-(sin 2x)/4 + (3 × cos 2x)/4} = sin x × {-sin 2x + 3 × cos 2x} = sin x × {-sin 2x + 3 × (1 – sin 2x)} = sin x × {-sin 2x + 3 – 3 × sin 2x} = sin x × {3 – 4 × sin 2x} = 3 × sin x – 4 sin 3x = sin 3x So, 4 × sin x × sin(x + π/3) × sin(x + 2π/3) = sin 3x

Question 19. If tan A – tan B = x and cot B – cot A = y, then the value of cot (A – B) is (a) x + y (b) 1/x + y (c) x + 1/y (d) 1/x + 1/y

Answer: (d) 1/x + 1/y Hint: Given, tan A – tan B = x ……………. 1 and cot B – cot A = y ……………. 2 From equation, 1/cot A – 1/cot B = x ⇒ (cot B – cot A)/(cot A × cot B) = x ⇒ y/(cot A × cot B) = x {from equation 2} ⇒ y = x × (cot A × cot B) ⇒ cot A × cot B = y/x Now, cot (A – B) = (cot A × cot B + 1)/(cot B – cot A) ⇒ cot (A – B) = (y/x + 1)/y ⇒ cot (A – B) = (y/x) × (1/y) + 1/y ⇒ cot (A – B) = 1/x + 1/y

Question 20. The value of (sin 7x + sin 5x) /(cos 7x + cos 5x) + (sin 9x + sin 3x) / (cos 9x + cos 3x) is (a) tan 6x (b) 2 tan 6x (c) 3 tan 6x (d) 4 tan 6x

Answer: (b) 2 tan 6x Hint: Given, (sin 7x + sin 5x) /(cos 7x + cos 5x) + (sin 9x + sin 3x) / (cos 9x + cos 3x) ⇒ [{2 × sin(7x+5x)/2 × cos(7x-5x)/2}/{2 × cos(7x+5x)/2 × cos(7x-5x)/2}] + [{2 × sin(9x+3x)/2 × cos(9x-3x)/2}/{2 × cos(9x+3x)/2 × cos(9x-3x)/2}] ⇒ [{2 × sin 6x × cosx}/{2 × cos 6x × cosx}] + [{2 × sin 6x × cosx}/{2 × cos 6x × cosx}] ⇒ (sin 6x/cos 6x) + (sin 6x/cos 6x) ⇒ tan 6x + tan 6x ⇒ 2 tan 6x

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CBSE Class 11 Term-1 Maths 2021-22: Topic-Wise MCQs, Case-Based & Assertion/Reason

CBSE Class 11 Term-1 Maths 2021-22: Topic-Wise MCQs, Case-Based & Assertion/Reason

In this Post I Have Provided Mathematics  Class 11 Chapter-Wise MCQs, Case-Based Question And Assertion/Reason For Term-1 Session 2021-22 With Solutions. CBSE has Recently Included These Types Of MCQs  And Assertion/Reason For Term-1 Exam 2020.   Every Student Knows these types of Questions Are Very Important For Their Term-1 Examination.

Any student who want to download these MCQs, then they have to click on given respective download links and it will be automatically download in your Google Drive.

CBSE Class 11th  Physics  Term-1 2021-22: Topic-Wise MCQs, Case-Based & Assertion/Reason

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Given Below Are The Class 11 Mathematics Chapter Name with Respective Download Links Containing Study Material:

CBSE Sample Papers For Class 11 Mathematics

CBSE sample paper for class  11th Maths  is an important tools to analyse itself. Before going to final examinations every student try to solve different types of sample question paper. It will enhance your knowledge and also provide to increase mental ability. Sample question paper are the best resources for the student to prepare for their Board examinations.

These sample question papers are very helpful to the student to get an entire experience before attempting the Board examinations papers. If students solve different types of sample question papers then they get the good confidence about the appropriate answer and they make good score in their Board examinations.

As every student know CBSE always change their exam pattern and also the model paper for their Board examinations that is why it is very important to all the student to understand the Model paper pattern before going to the final examinations.

In this articles I have provided different types of question papers on the basis of CBSE latest exam pattern which will definitely help to the student for the good preparation for their examinations.

In every year before the Board examination, CBSE provide model test paper to the student, on the basis of these model test papers I have prepared sample question papers for class  11th Maths  which will help to the student to understand the concept of model test paper and also it will help to the student for good preparation for their board examination.

CBSE Sample Paper:

Maths And Physics With Pandey Sir  is a website which provide free study material and notes to every students who are preparing for their Board examinations. By going in Level Section you can select CBSE sample papers. In this section, I have provided CBSE sample paper for class 9th to 12th.

Also if you are preparing for any competitive exams, they can find out study material and important notes in Level Section. In this website I have also provided most important MCQs questions, Assertion/Reason based questions and case study based questions for Board examinations or any other competitive examinations.

Benefits Of Solving CBSE Sample Papers for Examination:

Time Management:-

With the help of CBSE sample question papers you can make a proper strategy for their examinations also by practicing sample question papers you can manage your time during the Board examinations.

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After practicing different types of sample question papers you can understand on which section you have spend more time and which section is easy and which section is hard. In this way you can make a proper examination strategy.

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Self evaluation is very important to every student because with the help of self evaluation, student can understand their weak and strong point, which is very important for any Board examinations. After practicing more and more simple paper you can evaluate himself.

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As every student know there are many questions in their question paper which are very easy  but student make mistakes in this types of question. By solving different types of sample question papers you can reduce their silly mistakes.

Advantages Of CBSE Sample Question Papers:

* To get a good exam ideas about your examination paper patterns

* Student can easily access to the question papers

* It is very helpful to know about the mark distribution system.

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* To know about the pattern of CBSE

* Help to adjust a proper time management

* Available in PDF format

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Important Questions for CBSE Class 11 Maths Chapter 3 Trigonometric Functions 2024-25

• Class 11 Important Question
• Chapter 3: Trigonometric Functions

CBSE Class 11 Maths Chapter-3 Important Questions - Free PDF Download

Trigonometric functions class 11 important questions have been prepared for students of class 11 to help them score better marks in the examination. The complete topic of trigonometric functions is designed by the subject experts following the latest guidelines of CBSE . The class 11th maths trigonometric functions important questions feature the step by step solutions for easy to difficult questions according to the understanding level of the students. Moreover, students can download the PDF study materials from Vedantu that will help them to take a ready reference of the chapters and questions whenever they need. Based on the important questions, students can also make vital subject notes and mark them for a quick revision. As the students solve these important questions, they will easily develop a command over the topic of trigonometric functions. 

Download CBSE Class 11 Maths Important Questions 2024-25 PDF

Also, check CBSE Class 11 Maths Important Questions for other chapters:

Study Important Questions for Class 11 Mathematics Chapter 3 – Trigonometric Functions

1 Mark Questions

1. Find the radian measure corresponding to  $ 5{}^\circ \text{ }37'\text{ }30'' $  

Ans-  

Converting the given value to a pure degree form

 $ {{5}^{\circ }}37'30''={{5}^{\circ }}37'\left( \dfrac{30}{60} \right)' $ 

 $ \Rightarrow {{5}^{\circ }}37'60''={{5}^{\circ }}\left( \dfrac{75}{2} \right)' $ 

 $ \Rightarrow {{5}^{\circ }}37'60''={{5}^{\circ }}{{\left( \dfrac{75}{2\left( 60 \right)} \right)}^{\circ }} $ 

 $ \Rightarrow {{5}^{\circ }}37'60''={{\left( \dfrac{45}{8} \right)}^{\circ }} $ 

Degree to Radian Conversion

 $ \left( \dfrac{45}{8} \right)\left( \dfrac{\pi }{180} \right)=\dfrac{\pi }{32}\text{rad} $ 

2. Find degree measure corresponding to  $ {{\left( \dfrac{\pi }{16} \right)}^{c}} $  

Converting the given value from radian to degree form

$ \dfrac{\pi }{16}\times \dfrac{180}{\pi }={{\left( \dfrac{45}{4} \right)}^{\circ }} $ 

Simplify degree form

$ {{\left( \dfrac{45}{4} \right)}^{\circ }}={{11}^{\circ }}15' $ 

3. Find the length of an arc of a circle of radius  $ 5cm $  subtending a central angle measuring  $ 15{}^\circ  $  

The arc of a circle with a radius of  $ 5\,\text{cm} $  with a central angle of  $ {{15}^{\circ }} $  should be of the length  $ \dfrac{5\pi }{12}cm $  using the formula  $ \text{Arc}\,\text{=}\,\pi \times \left( \theta  \right) $ .

4. Find the value of  $ \dfrac{19\pi }{3} $  

We have  $ \tan \dfrac{19\pi }{3} $ 

 $ \tan \dfrac{19\pi }{3}=\tan \left( 6\dfrac{\pi }{3} \right) $ 

                   $ =\tan \left( 6\pi +\dfrac{\pi }{3} \right) $ 

                   $ =\tan \left( 3\times 2\pi +\dfrac{\pi }{3} \right) $ 

                   $ =\tan \left( \dfrac{\pi }{3} \right) $ 

                   $ =\sqrt{3} $ 

5. Find the value of  $ \sin \left( -1125{}^\circ  \right) $  

We have  $ \sin \left( -{{1125}^{\circ }} \right) $ 

 $ \sin \left( -\dfrac{1125}{360}\times {{360}^{\circ }} \right) $ 

 $ =-\sin \left( \left( 3+\dfrac{45}{360} \right)\times {{360}^{\circ }} \right) $ 

 $ =-\sin \left( {{45}^{\circ }} \right) $ 

 $ =-\dfrac{1}{\sqrt{2}} $ 

6. Find the value of  $ \tan \left( {{15}^{\circ }} \right) $  

We have  $ \tan {{15}^{\circ }} $ 

 $ \tan {{15}^{\circ }}=\tan \left( {{60}^{\circ }}-{{45}^{\circ }} \right) $ 

                 $ =\dfrac{\tan {{60}^{\circ }}-\tan {{45}^{\circ }}}{1+\tan {{60}^{\circ }}\times \tan {{45}^{\circ }}} $ 

                 $ =\dfrac{\sqrt{3}-1}{\sqrt{3}+1} $ 

7. If  $ \sin A=\dfrac{3}{5} $   and  $ \dfrac{\pi }{2}<A< $  find $ \cos A $  

The condition  $ \dfrac{\pi }{2}<A $  denotes that we need to take into account for the second quadrant , hence the cosine value will be negative . 

Therefore, 

 $ \cos A=\dfrac{-4}{5} $ 

8. If  $ \tan A=\dfrac{a}{a+1} $   and  $ \tan B=\dfrac{1}{2a+1} $   then find the value of  $ A+B $  

 $ \tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B} $ 

                           $ =\dfrac{\dfrac{a}{a+1}+\dfrac{1}{2a+1}}{1-\dfrac{a}{a+1}\cdot \dfrac{1}{2a+1}} $ 

                          $ =\dfrac{\dfrac{2{{a}^{2}}+2a+1}{\left( a+1 \right)\left( 2a+1 \right)}}{\dfrac{\left( a+1 \right)\left( 2a+1 \right)-a}{\left( a+1 \right)\left( 2a+1 \right)}} $ 

                          $ =1 $ 

Which can only be possible if  $ A+B={{45}^{\circ }} $ .

9. Express  $ \sin 12\theta +\sin 4\theta  $   as the product of sines and cosine

Using the trigonometric difference formula, we get

 $ \sin 12\theta +\sin 4\theta =\sin \left( 8\theta +4\theta  \right)+\sin \left( 8\theta -4\theta  \right) $  

                                    $ =2\sin 8\theta \cos 4\theta  $ 

10. Express  $ 2\cos 4x\sin 2x $   as an algebraic sum of sines or cosine.

 $ 2\cos 4x\sin 2x=\sin \left( 2x+4x \right)+\sin \left( 2x-4x \right) $ 

                                 $ =\sin 6x+\sin \left( -2x \right) $ 

                                  $ =\sin 6x-\sin 2x $ 

11. Write the range of  $ \cos \theta  $  

The cosine function is a periodic function with a domain of  $ \mathbb{R} $  and a range of  $ \left[ -1,1 \right] $ .

12. What is domain of  $ \sec \theta  $  

The secant function is the reciprocal of the cosine function , it has a domain of  $ \mathbb{R}-\left\{ (2n+1)\dfrac{\pi }{2};n\in \mathbb{Z} \right\} $  because those are the points where the cosine function equates to  $ 0 $ .

13. Find the principal solution of  $ \cot x=3 $  

The principal solution of  $ \cot x=3 $  is for the following input values  $ x=\dfrac{5\pi }{6},\dfrac{11\pi }{6} $ .

14. Write the general solution of  $ \cos \theta =0 $  

The general solution for the equation  $ \cos \theta =0 $  is  $ \theta =(2n+1)\dfrac{\pi }{2},n\in \mathbb{Z} $ .

15. If  $ \sin x=\dfrac{\sqrt{5}}{3} $ and  $ 0\text{ }<\text{ }x\text{ }<\dfrac{\pi }{2} $ find the value of $ \cos 2x $  

We know that  $ \cos 2x=1-{{\sin }^{2}}x $ 

 $ \cos 2x=1-2{{\left( \dfrac{\sqrt{5}}{3} \right)}^{2}} $ 

                $ =1-2\times \dfrac{5}{9} $ 

                $ =-\dfrac{1}{9} $ 

16. If  $ \cos x=-\dfrac{1}{3} $   and  $ x $   lies in quadrant  $ \text{III} $ , find the value of  $ \sin \dfrac{x}{2} $  

We know that  $ \cos 2x=1-2{{\sin }^{2}}x $ 

 $ \cos \left( 2\left( \dfrac{x}{2} \right) \right)=1-2{{\sin }^{2}}\left( \dfrac{x}{2} \right) $ 

 $ \Rightarrow -\dfrac{1}{3}=1-2{{\sin }^{2}}\dfrac{x}{2} $ 

 $ \Rightarrow 2{{\sin }^{2}}\dfrac{x}{2}=1+\dfrac{1}{3} $ 

 $ \Rightarrow {{\sin }^{2}}\dfrac{x}{2}=\dfrac{2}{3} $ 

 $ \Rightarrow \sin \dfrac{x}{2}=\pm \sqrt{\dfrac{2}{3}} $ 

 $ \Rightarrow \sin \dfrac{x}{2}=\sqrt{\dfrac{2}{3}}\,\,\,\,\,\left[ \text{2nd Quadrant} \right] $ 

17. Convert into radian measures  $ -47{}^\circ 30' $  

Convert into pure degree form and then convert to radian

 $ -47{}^\circ 30'=-{{\left( 47+\dfrac{30}{60} \right)}^{{}^\circ }} $ 

                   $ =-{{\left( 47+\dfrac{1}{2} \right)}^{{}^\circ }} $ 

                   $ =-\left( \dfrac{95}{2}\times \dfrac{\pi }{180} \right)\text{rad} $ 

                   $ =-\dfrac{19\pi }{72}\text{rad} $ 

18. Evaluate  $ \tan 75{}^\circ  $  

Use the trigonometric addition formula for the tangent function

 $ \tan {{75}^{\circ }}=\tan ({{45}^{\circ }}+{{30}^{\circ }}) $ 

                 $ =\dfrac{\tan {{45}^{\circ }}+\tan {{30}^{\circ }}}{1-\tan {{45}^{\circ }}\tan {{30}^{\circ }}} $ 

                 $ =\dfrac{\sqrt{3}+1}{\sqrt{3}-1} $ 

19. Prove that  $ \sin (40+\theta )\cdot \cos (10+\theta )-\cos (40+\theta )\cdot \sin (10+\theta )=\dfrac{1}{2} $  

Ans-  

Let us take the left-hand side of the equation and make some manipulations.

We know,  $ \sin \left( a-b \right)=\sin a\cos b-\cos a\sin b $ 

 $ \text{L}\text{.H}\text{.S}=\sin (40+\theta )\cos (10+\theta )-\cos (40+\theta )\sin (10+\theta ) $ 

              $ =\sin \left[ 40+\theta -10-\theta  \right]=\sin 30 $ 

              $ =\dfrac{1}{2} $ 

20. Find the principal solution of the eq.  $ \sin x=\dfrac{\sqrt{3}}{2} $  

The principal solution of  $ \sin x=\dfrac{\sqrt{3}}{2} $  is the input values of  $ x=\dfrac{\pi }{3},\dfrac{2\pi }{3} $ 

21. Prove that  $ \cos \left( \dfrac{\pi }{4}+x \right)+\cos \left( \dfrac{\pi }{4}-x \right)=\sqrt{2}\cos x $ 

Let us start with the left-hand side and use the trigonometric differences formula for the cosine function

 $ \text{L}\text{.H}\text{.S}=\cos \left( \dfrac{\pi }{4}+x \right)+\cos \left( \dfrac{\pi }{4}-x \right) $ 

               $ =2\cos \dfrac{\pi }{4}\cos x $ 

               $ =2\left( \dfrac{1}{\sqrt{2}} \right)\cos x $ 

               $ =\sqrt{2}\cos x $ 

               $ =\text{R}\text{.H}\text{.S} $ 

22. Convert into radian measures  $ -37{}^\circ 30' $  

Convert into pure degree form and then convert from degree to radian

 $ -37{}^\circ 30'={{\left( 37+\dfrac{30}{60} \right)}^{{}^\circ }} $ 

                 $ =-{{\left( \dfrac{75}{2} \right)}^{{}^\circ }} $ 

                 $ =-\dfrac{75}{2}\times \dfrac{\pi }{180}\text{rad} $ 

                 $ =-\dfrac{5\pi }{24}\text{rad} $ 

 $ Sin\text{ }\left( n+1 \right)\text{ }x\text{ }Sin\text{ }\left( n+2 \right)\text{ }x\text{ }+\text{ }Cos\text{ }\left( n+1 \right)\text{ }x.\text{ }Cos\text{ }\left( n+2 \right)\text{ }x\text{ }=\text{ }Cos\text{ }x $   

 $ \text{L}\text{.H}\text{.S}\,\text{. = sin}\left( n+1 \right)x\sin \left( n+2 \right)x+\cos \left( n+1 \right)x\cos \left( n+2 \right)x $ 

                 $ =\cos \left\{ \left( n+1 \right)x-\left( n+2 \right)x \right\} $ 

                 $ =\cos \left( nx+x-n-2x \right) $ 

                 $ =\cos \left( -x \right) $ 

                 $ =\cos \left( x \right) $ 

Find the value of  $ \operatorname{Sin}\dfrac{31\pi }{3} $  

We have  $ \sin \dfrac{31\pi }{3} $ 

 $ \operatorname{Sin}\dfrac{31\pi }{3}=\operatorname{Sin}\left( 10\pi +\dfrac{\pi }{3} \right) $ 

                   $ =\operatorname{Sin}\left( 2\pi \times 5+\dfrac{\pi }{3} \right)\,\,\,\,\,\,\left[ \text{Periodic Function} \right] $ 

                   $ =\operatorname{Sin}\dfrac{\pi }{3} $ 

                   $ =\dfrac{\sqrt{3}}{2} $ 

Find the principal solution of the eq.  $ \tan x=-\dfrac{1}{\sqrt{3}} $ . 

The principal solution of the equation  $ \tan x=-\dfrac{1}{\sqrt{3}} $  will be the input values of  $ x=\dfrac{5\pi }{6},\dfrac{11\pi }{6} $ 

Convert into radian measures $ 5{}^\circ \text{ }37'\text{ }30'' $  

  Prove  $ Cos70{}^\circ .\text{ }Cos10{}^\circ +\text{ }Sin70{}^\circ .\text{ }Sin10{}^\circ =\dfrac{1}{2} $  

Starting with the left-hand side and using the trigonometric differences formula for the cosine function .

 $ \text{L}\text{.H}\text{.S}=\text{cos}\left( {{70}^{\circ }}{{10}^{\circ }} \right) $ 

               $ =\cos {{60}^{\circ }} $ 

               $ =\dfrac{1}{2} $ 

  Evaluate   $ 2\operatorname{Sin}\dfrac{\pi }{12} $  

Use the trigonometric difference formula for the sine function and expand

 $ 2\sin \dfrac{\pi }{12}=2\sin \left[ \dfrac{\pi }{4}-\dfrac{\pi }{6} \right] $ 

                   $ =2\left[ \sin \dfrac{\pi }{4}\cos \dfrac{\pi }{6}-\cos \dfrac{\pi }{4}\sin \dfrac{\pi }{6} \right] $ 

                   $ =2\left[ \dfrac{1}{\sqrt{2}}\times \dfrac{\sqrt{3}}{2}-\dfrac{1}{\sqrt{2}}\times \dfrac{1}{2} \right] $ 

                    $ =\dfrac{\sqrt{3}-1}{\sqrt{2}} $ 

 Find the solution of  $ \operatorname{Sin}x=-\dfrac{\sqrt{3}}{2} $  

We are required to find the general solution for the equation  $ \sin x=-\dfrac{\sqrt{3}}{2} $ 

 $ \operatorname{Sin}x=-\dfrac{\sqrt{3}}{2} $ 

 $ \Rightarrow \operatorname{Sin}x=\operatorname{Sin}\left( \pi +\dfrac{\pi }{3} \right) $ 

 $ \Rightarrow \operatorname{Sin}x=\operatorname{Sin}\dfrac{4\pi }{3} $ 

 $ \operatorname{Sin}\theta =\operatorname{Sin}\alpha  $ 

 $ \theta =n\pi +{{(-1)}^{n}}\cdot \alpha  $ 

 $ x=n\pi +{{(-1)}^{n}}\cdot \dfrac{4\pi }{3} $ 

 Prove that  $ \dfrac{\operatorname{Cos}9{}^\circ -\operatorname{Sin}9{}^\circ }{\operatorname{Cos}9{}^\circ +\operatorname{Sin}9{}^\circ }=\tan 36{}^\circ  $  

Let us start with the right-hand side and use the trigonometric differences formula for the tangent function.

 $ \text{R}\text{.H}\text{.S}=\tan 36{}^\circ  $ 

                $ =\tan \left( {{45}^{\circ }}-{{9}^{\circ }} \right) $ 

                $ =\dfrac{\tan {{45}^{\circ }}-\tan {{9}^{\circ }}}{1+\tan {{45}^{\circ }}\tan {{9}^{\circ }}} $ 

                $ =\dfrac{1-\tan {{9}^{\circ }}}{1+\tan {{9}^{\circ }}} $ 

                $ =\dfrac{\cos {{9}^{\circ }}-\sin {{9}^{\circ }}}{\cos {{9}^{\circ }}+\sin {{9}^{\circ }}} $ 

                $ =\text{L}\text{.H}\text{.S}\text{.} $ 

 Find the value of  $ \tan \dfrac{19\pi }{3} $  

We have  $ \tan \left( \dfrac{19\pi }{3} \right) $ 

 $ \tan \dfrac{19\pi }{3}=\tan \left( 6\pi -\dfrac{\pi }{3} \right) $ 

                  $ =\tan \left[ 3\times 2\pi +\dfrac{\pi }{3} \right]\,\,\,\,\,\,\,\,\,\,\left[ \text{Periodic Function} \right] $ 

                  $ =\tan \dfrac{\pi }{3} $ 

                  $ =\sqrt{3} $ 

  Prove   $ \operatorname{Cos}4x=1-8{{\operatorname{Sin}}^{2}}x.{{\operatorname{Cos}}^{2}}x $ 

Starting with the left-hand side and using the trigonometric addition formula, $ \cos 2x=1-2{{\sin }^{2}}x $ 

 $ \text{L}\text{.H}\text{.S}=\operatorname{Cos}4x $ 

               $ =1-2{{\operatorname{Sin}}^{2}}2x $ 

               $ =1-2{{(\operatorname{Sin}2x)}^{2}} $ 

               $ =1-2{{(2\operatorname{Sin}x.\operatorname{Cos}x)}^{2}} $ 

               $ =1-2(4{{\operatorname{Sin}}^{2}}x.{{\operatorname{Cos}}^{2}}x) $ 

               $ =1-8{{\operatorname{Sin}}^{2}}x.{{\operatorname{Cos}}^{2}}x $ 

 Prove  $ \dfrac{\operatorname{Cos}(\pi +x).\operatorname{Cos}(-x)}{\operatorname{Sin}(\pi -x).\operatorname{Cos}\left( \dfrac{\pi }{2}+x \right)}=Co{{t}^{2}}x $  

Starting with the left-hand side and using the trigonometric periodic identities, we obtain the following

 $ \text{L}\text{.H}\text{.S}\text{.}=\dfrac{\cos \left( \pi +x \right)\cos \left( -x \right)}{\sin \left( \pi -x \right)\cos \left( \dfrac{\pi }{2}+x \right)} $ 

                 $ =\dfrac{-\cos x\cos x}{-\sin x\sin x} $ 

                 $ ={{\cot }^{2}}x $ 

                 $ =\text{R}\text{.H}\text{.S}\text{.} $ 

 Prove that  $ \tan {{56}^{\circ }}=\dfrac{\operatorname{Cos}{{11}^{\circ }}+\operatorname{Sin}{{11}^{\circ }}}{\operatorname{Cos}{{11}^{\circ }}-\operatorname{Sin}{{11}^{\circ }}} $  

Starting with the left-hand side and using the trigonometric addition formula for the tangent function, we obtain

 $ \text{L}\text{.H}\text{.S}\text{.}=\tan {{56}^{\circ }} $ 

                 $ =\tan ({{45}^{\circ }}+{{11}^{\circ }}) $ 

                 $ =\dfrac{\tan {{45}^{\circ }}+\tan {{11}^{\circ }}}{1-\tan {{45}^{\circ }}\cdot \tan {{11}^{\circ }}} $ 

                 $ =\dfrac{1+\tan {{11}^{\circ }}}{1-\tan {{11}^{\circ }}} $ 

                $ =\dfrac{\cos {{11}^{\circ }}+\sin {{11}^{\circ }}}{\cos {{11}^{\circ }}-\sin {{11}^{\circ }}} $ 

                $ =\text{R}\text{.H}\text{.S}\text{.} $ 

  Prove that  $ \operatorname{Cos}{{105}^{\circ }}+\operatorname{Cos}{{15}^{\circ }}=\operatorname{Sin}{{75}^{\circ }}-\operatorname{Sin}{{15}^{\circ }} $ 

Starting with the left-hand side and using the trigonometric difference formula for the cosine function, we obtain

 $ \text{L}\text{.H}\text{.S}\text{.}=\operatorname{Cos}{{105}^{\circ }}+\operatorname{Cos}{{15}^{\circ }} $ 

                 $ =\operatorname{Cos}({{90}^{\circ }}+{{15}^{\circ }})+\operatorname{Cos}({{90}^{\circ }}-{{75}^{\circ }}) $ 

                 $ =-\operatorname{Sin}{{15}^{\circ }}+\operatorname{Sin}{{75}^{\circ }} $ 

                 $ =\operatorname{Sin}{{75}^{\circ }}-\operatorname{Sin}{{15}^{\circ }} $ 

  Find the value of   $ \operatorname{Cos}(-{{1710}^{\circ }}) $ 

We have  $ \cos \left( -{{1710}^{\circ }} \right) $ . We also know  $ \cos \left( -x \right)=\cos x $ 

 $ \operatorname{Cos}(-{{1710}^{\circ }})=\operatorname{Cos}(1800-90) $ 

                               $ =\operatorname{Cos}\left[ 5\times 360+90 \right] $ 

                               $ =\operatorname{Cos}\dfrac{\pi }{2} $ 

                               $ =0 $ 

   A wheel makes  $ 360 $  revolutions in  $ 1 $  minute. Through how many radians does it turn in  $ 1 $  second.

 $ \text{Number of revolutions made in 60s}=360 $ 

 $ \text{Number of revolutions made in 1s}=\dfrac{360}{60} $ 

 $ \text{Angle moved in 6 revolutions}=2\pi \times 6 $ 

 $ =12\pi  $ 

Prove that  $ {{\operatorname{Sin}}^{2}}6x-{{\operatorname{Sin}}^{2}}4x=\operatorname{Sin}2x.\operatorname{Sin}10x $ 

Starting with the left-hand side and using the trigonometric addition formula for the sine function, we obtain

 $ \text{L}\text{.H}\text{.S}\text{.}={{\operatorname{Sin}}^{2}}6x-{{\operatorname{Sin}}^{2}}4x $ 

                $ =\sin \left( 6x+4x \right)\sin \left( 6x-4x \right) $ 

                $ =\sin 10x\sin 2x $ 

Prove that  $ \dfrac{\tan 69+\tan 66}{1\tan 69.\tan 66}=-1 $ 

Starting with the left-hand side and using the trigonometric difference identity for the tangent function, we obtain

 $ \text{L}\text{.H}\text{.S}\text{.}=\dfrac{\tan {{69}^{\circ }}+\tan {{66}^{\circ }}}{1-\tan {{69}^{\circ }}\tan {{66}^{\circ }}} $ 

                 $ =\tan ({{69}^{\circ }}+{{66}^{\circ }}) $ 

                 $ =\tan \left( {{135}^{\circ }} \right) $ 

                 $ =\tan \left( {{90}^{\circ }}+{{45}^{\circ }} \right) $ 

                $ =-1 $ 

 Prove that  $\dfrac{\operatorname{Sin}x}{1+\operatorname{Cos}x}=\tan 

\dfrac{x}{2} $  

Starting with the left-hand side and using the trigonometric addition identities for the sine and cosine function, we obtain

 $ \text{L}\text{.H}\text{.S}\text{.}=\dfrac{\sin x}{1+\cos x} $ 

                 $ =\dfrac{2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}{2{{\cos }^{2}}\dfrac{x}{2}} $ 

                 $ =\dfrac{\sin \dfrac{x}{2}}{\cos \dfrac{x}{2}} $ 

                 $ =\tan \dfrac{x}{2} $ 

4 Marks Questions

Prove the following identities

1. The minute hand of a watch is  $ 1.5cm $   long. How far does its tip move in  $ 40 $  minute? 

Analysing the given information, we have 

 $ r=1.5cm $ 

 $ \text{Angle made in }60\min ={{360}^{\circ }} $ 

 $ \text{Angle made in 1min}={{6}^{\circ }} $ 

 $ \text{Angle made in 40min}={{6}^{\circ }}\times {{40}^{\circ }}={{240}^{\circ }} $ 

Calculating the arc distance

              $ \theta =\dfrac{l}{r} $ 

 $ 240\times \dfrac{\pi }{180}=\dfrac{l}{1.5} $ 

      $ 2\times 3.14=l $ 

              $ 6.28=l $ 

                      $ l=6.28cm $ 

2. Show that  $ tan\text{ }3x.\text{ }tan\text{ }2x.\text{ }tan\text{ }x\text{ }=\text{ }tan\text{ }3x\text{ }\text{ }tan\text{ }2x\text{ }\text{ }tan\text{ }x $  

Let us start with  $ \tan 3x $  and we know  $ 3x=2x+x $ 

 $ \tan 3x=\tan (2x+x) $ 

 $ \dfrac{\tan 3x}{1}=\dfrac{\tan 2x+\tan x}{1-\tan 2x.\tan x} $ 

 $ \tan 3x(1-\tan 2x.\tan x)=\tan 2x+\tan x $ 

 $ \tan 3x-\tan 3x.\tan 2x.\tan x=\tan 2x+\tan x $ 

 $ \tan 3x.\tan 2x.\tan x=\tan 3x-\tan 2x-\tan x $ 

3. Find the value of  $ \tan \dfrac{\pi }{8} $  

We know that

 $ \tan 2x=\dfrac{2\tan x}{1-{{\tan }^{2}}x} $ 

Therefore, we have

    $ \tan \left( 2\dfrac{\pi }{8} \right)=\dfrac{2\tan \dfrac{\pi }{8}}{1-{{\tan }^{2}}\dfrac{\pi }{8}} $ 

 $ \Rightarrow 1=\dfrac{2\tan \dfrac{\pi }{8}}{1-{{\tan }^{2}}\dfrac{\pi }{8}} $ 

Put  $ \tan \dfrac{\pi }{8}=x $ 

 $ 1=\dfrac{2x}{1-{{x}^{2}}} $ 

 $ \Rightarrow 2x=1-{{x}^{2}} $ 

 $ \Rightarrow x=\dfrac{-1\pm \sqrt{2}}{1} $ 

Since,  $ \dfrac{\pi }{8} $  lies in the first quadrant, the value must be positive, hence

$ \tan \dfrac{\pi }{8}=\sqrt{2}-1 $ 

4. Prove that  $ \dfrac{\operatorname{Sin}(x+y)}{\operatorname{Sin}(x-y)}=\dfrac{\tan x+\tan y}{\tan x-\tan y} $  

Starting with the left-hand side and using the trigonometric difference formula for the sine function, we get

 $ \text{L}\text{.H}\text{.S}\text{.}=\dfrac{\operatorname{Sin}(x+y)}{\operatorname{Sin}(x-y)} $ 

                $ =\dfrac{\operatorname{Sin}x.\operatorname{Cos}y+\operatorname{Cos}x.\operatorname{Sin}y}{\operatorname{Sin}x.\operatorname{Cos}y-\operatorname{Cos}x.\operatorname{Sin}y} $ 

Dividing numerator and denominator by  $ \operatorname{Cos}x.\operatorname{Cos}y $ 

           $ =\dfrac{\tan x+\tan y}{\tan x-\tan y} $ 

               $ =\text{R}\text{.H}\text{.S}\text{.} $ 

5. If in two circles, arcs of the same length subtend angles  $ {{60}^{\circ }} $  and  $ {{75}^{\circ }} $  at the center find the ratio of their radii. 

We know that the length of the arc and its subtended angle is related using the following formula

 $ \theta =\dfrac{1}{{{r}_{1}}} $ 

 $ 60\times \dfrac{\pi }{18}=\dfrac{1}{{{r}_{1}}} $ 

 $ {{r}_{1}}=\dfrac{3l}{\pi } $      ….. $ (1) $ 

 $ \theta =\dfrac{1}{{{r}_{2}}} $ 

 $ 75\times \dfrac{\pi }{18}=\dfrac{1}{{{r}_{2}}} $ 

 $ {{r}_{2}}=\dfrac{12l}{5\pi } $     ….. $ (2) $ 

 $ (1)\div (2) $ 

 $ \dfrac{{{r}_{1}}}{{{r}_{2}}}=\dfrac{\dfrac{3l}{\pi }}{\dfrac{12l}{5\pi }} $ 

 $ =\dfrac{31}{\pi }\times \dfrac{5\pi }{12l} $ 

 $ =5:4 $ 

6. Prove that  $ \operatorname{Cos}6x=32{{\operatorname{Cos}}^{2}}x-48{{\operatorname{Cos}}^{4}}x+18{{\operatorname{Cos}}^{2}}x-1 $  

Starting with the left-hand side and using the trigonometric identities for the cosine function, we obtain

 $ \text{L}\text{.H}\text{.S}\text{.}=\operatorname{Cos}6x $ 

                 $ =\operatorname{Cos}2(3x)=2{{\operatorname{Cos}}^{2}}3x-1 $ 

                 $ =\operatorname{Cos}2(3x) $ 

                 $ =2{{(4co{{s}^{3}}x-3\cos x)}^{2}}-1 $ 

                 $ =2\left[ 16{{\operatorname{Cos}}^{6}}x+9{{\operatorname{Cos}}^{2}}x-24{{\operatorname{Cos}}^{4}}x \right]-1 $ 

                  $ =32{{\operatorname{Cos}}^{6}}x+18{{\operatorname{Cos}}^{2}}x-48{{\operatorname{Cos}}^{4}}x-1 $ 

               $ =32{{\operatorname{Cos}}^{6}}x-48{{\operatorname{Cos}}^{4}}x+18{{\operatorname{Cos}}^{2}}x1 $ 

                  $ =\text{R}\text{.H}\text{.S}\text{.} $ 

7. Solve  $ \operatorname{Sin}2x-\operatorname{Sin}4x+\operatorname{Sin}6x=0 $  

Starting with the left-hand side and using the trigonometric addition identity for the sine function, we obtain

 $ \text{L}\text{.H}\text{.S}\text{.}=\operatorname{Sin}6x+\operatorname{Sin}2x-\operatorname{Sin}4x $ 

                 $ =2\sin \left( \dfrac{6x+2x}{2} \right)\cos \left( \dfrac{6x-2x}{2} \right)-\sin 4x $ 

                 $ =\sin 4x\left( 2\cos 2x-1 \right) $ 

                 $ =0 $ 

 $ \sin 4x=0 $ 

      $ 4x=n\pi  $ 

        $ x=\dfrac{n\pi }{4} $ 

 $ 2\cos 2x-1=0 $ 

        $ \cos 2x=\cos \dfrac{\pi }{3} $ 

             $ 2x=2n\pi \pm \dfrac{\pi }{3} $ 

              $ x=n\pi \pm \dfrac{\pi }{6} $ 

8. In a circle of diameter  $ 40cm $ , the length of a chord is  $ 20cm $ . Find the length of the minor area of the chord. 

 $ \theta =\dfrac{l}{r} $ 

 $ \Rightarrow 60\times \dfrac{\pi }{180}=\dfrac{l}{20} $ 

 $ \Rightarrow l=\dfrac{20\pi }{3}\text{cm/s} $ 

9. Prove that  $ \tan 4x=\dfrac{4\tan x(1-{{\tan }^{2}}x)}{1-6{{\tan }^{2}}x+{{\tan }^{4}}x} $  

Starting with the left-hand side and using the trigonometric addition identities for the tangent function, we obtain

 $ \text{L}\text{.H}\text{.S}\text{.}=\tan 4x $ 

                 $ =\dfrac{2\tan 2x}{1-{{\tan }^{2}}2x} $ 

                 $ =\dfrac{2.\dfrac{2\tan 2x}{1-{{\tan }^{2}}2x}}{1-{{\left( \dfrac{2\tan 2x}{1-{{\tan }^{2}}2x} \right)}^{2}}} $ 

                $ =\dfrac{\dfrac{4\tan x}{1-{{\tan }^{2}}x}}{\dfrac{{{(1-{{\tan }^{2}}x)}^{2}}-4{{\tan }^{2}}x}{{{(1-{{\tan }^{2}}x)}^{2}}}} $ 

                 $ =\dfrac{4\tan x}{(1-{{\tan }^{2}}x)}\times \dfrac{(1-{{\tan }^{2}}x)}{1+{{\tan }^{4}}x-2{{\tan }^{2}}x-4{{\tan }^{2}}x} $ 

                 $ =\dfrac{4\tan x(1-{{\tan }^{2}}x)}{1-6{{\tan }^{2}}x+{{\tan }^{4}}x} $ 

10. Prove that  $ {{\left( Cosx+Cosy \right)}^{2}}+{{\left( SinxSiny \right)}^{2}}=4Co{{s}^{2}}\left( \dfrac{x+y}{2} \right) $  

Starting with the left-hand side and using the trigonometric addition identities for the cosine and sine function, we obtain

 $ \text{L}\text{.H}\text{.S}\text{.}={{\left( Cosx+Cosy \right)}^{2}}+{{\left( SinxSiny \right)}^{2}} $ 

                $ ={{\left( 2\operatorname{Cos}\dfrac{x+y}{2}.\operatorname{Cos}\dfrac{x-y}{2} \right)}^{2}}+{{\left( 2\operatorname{Cos}\left( \dfrac{x+y}{2} \right).\operatorname{Sin}\left( \dfrac{x-y}{2} \right) \right)}^{2}} $ 

                $ =4{{\operatorname{Cos}}^{2}}\dfrac{x+y}{2}.{{\operatorname{Cos}}^{2}}\left( \dfrac{x-y}{2} \right)+4{{\operatorname{Cos}}^{2}}\dfrac{x+y}{2}.{{\operatorname{Sin}}^{2}}\dfrac{x-y}{2} $ 

               $ =4{{\operatorname{Cos}}^{2}}\left( \dfrac{x+y}{2} \right)\left[ {{\operatorname{Cos}}^{2}}\dfrac{x-y}{2}+{{\operatorname{Sin}}^{2}}\dfrac{x-y}{2} \right] $ 

               $ =4{{\operatorname{Cos}}^{2}}\left( \dfrac{x+y}{2} \right) $ 

11. If  $ Cotx=-\dfrac{5}{12},x $  lies in second quadrant find the values of other five trigonometric functions 

 $ Cotx=-\dfrac{5}{12} $ 

Using some trigonometric identities, we obtain

 $ \tan x=-\dfrac{12}{5} $ 

 $ {{\operatorname{Sec}}^{2}}x=1+{{\tan }^{2}}x $ 

 $ \operatorname{Sec}x=\pm \dfrac{13}{5} $ 

Since  $ x $  lies in the second quadrant, the cosine value will be negative

 $ \operatorname{Sec}x=-\dfrac{13}{5} $ 

 $ \operatorname{Cos}x=-\dfrac{5}{13} $ 

 $ \operatorname{Sin}x=\tan x.\operatorname{Cos}x $ 

             $ =\dfrac{-12}{5}\times \left( \dfrac{-5}{13} \right) $ 

              $ =\dfrac{12}{13} $ 

 $ \operatorname{Csc}x=\dfrac{13}{12} $ 

12. Prove that  $ \dfrac{\operatorname{Sin}5x-2\operatorname{Sin}3x+\operatorname{Sin}x}{\operatorname{Cos}5x-\operatorname{Cos}x}=\tan x $  

Starting with the left-hand side and using the trigonometric difference identities for the sine function, we obtain

 $ \text{L}\text{.H}\text{.S}\text{.}=\dfrac{\operatorname{Sin}5x+\operatorname{Sin}x-2\operatorname{Sin}3x}{\operatorname{Cos}5x-\operatorname{Cos}x} $ 

                 $ =\dfrac{2\operatorname{Sin}3x.\operatorname{Cos}2x-2\operatorname{Sin}3x}{-2\operatorname{Sin}3x.\operatorname{Sin}2x} $ 

                 $ =\dfrac{2\operatorname{Sin}3x(\operatorname{Cos}2x-1)}{-2\operatorname{Sin}3x.\operatorname{Sin}2x} $ 

                 $ =\dfrac{-(1-\operatorname{Cos}2x)}{-\operatorname{Sin}2x} $ 

                 $ =\dfrac{2{{\operatorname{Sin}}^{2}}x}{2\operatorname{Sin}x.\operatorname{Cos}x} $ 

                 $ =\dfrac{\operatorname{Sin}x}{\operatorname{Cos}x} $ 

                 $ =\tan x $ 

13. Prove that  $ Sinx+Sin3x+Sin5x+Sin7x=4Cosx.Cos2x.Sin4x $  

Starting with the left-hand side and using the trigonometric addition identities for the sine function, we obtain

 $ \text{L}\text{.H}\text{.S}\text{.}=Sinx+Sin3x+Sin5x+Sin7x $ 

                 $ =\operatorname{Sin}x+\operatorname{Sin}7x+\operatorname{Sin}3x+\operatorname{Sin}5x $ 

                 $ =2\operatorname{Sin}\left( \dfrac{x+7x}{2} \right).\operatorname{Cos}\left( \dfrac{x-7x}{2} \right)+2\operatorname{Sin}\left( \dfrac{3x+5x}{2} \right)\operatorname{Cos}\left( \dfrac{3x-5x}{2} \right) $ 

                 $ =2\operatorname{Sin}4x.\operatorname{Cos}3x+2\operatorname{Sin}4x.\operatorname{Cos}x $ 

                 $ =2\operatorname{Sin}4x[\operatorname{Cos}3x+\operatorname{Cos}x] $ 

                 $ =2\operatorname{Sin}4x\left[ 2\operatorname{Cos}\left( \dfrac{3x+x}{2} \right).\operatorname{Cos}\left( \dfrac{3x-x}{2} \right) \right] $ 

                 $ =2\operatorname{Sin}4x[2\operatorname{Cos}2x.\operatorname{Cos}x] $ 

                 $ =4\operatorname{Cos}x.\operatorname{Cos}2x.\operatorname{Sin}4x $ 

14. Find the angle between the minute hand and hour hand of a clock when the time is  $ 7.20 $  

We know that the angle made by minute hand in  $ 15\min =15\times 6={{90}^{\circ }} $ 

We also know that the angle made by the hour hand in  $ 1hr={{30}^{\circ }} $ 

In  $ 60 $  minute  $ =\dfrac{30}{60} $ 

 $ =\dfrac{1}{2} $ 

 $ [\because  $ Angle Travelled by  $ hr $ hand in  $ 12hr={{360}^{\circ }}] $ 

In  $ 20 $ minutes  $ =\dfrac{1}{2}\times 20 $ 

                         $ ={{10}^{\circ }} $ 

Angle made  $ =90+10 $ 

                       $ ={{100}^{\circ }} $ 

15. Show that  $ \sqrt{2+\sqrt{2+2\operatorname{Cos}4\theta }}=2\operatorname{Cos}\theta  $  

Starting with the left-hand side and using the trigonometric addition identity for the cosine function, we obtain

 $ \text{L}\text{.H}\text{.S}\text{.}=\sqrt{2+\sqrt{2+2\operatorname{Cos}4\theta }} $ 

                 $ =\sqrt{2+\sqrt{2(1+\operatorname{Cos}4\theta )}} $ 

                 $ =\sqrt{2+\sqrt{2.2{{\operatorname{Cos}}^{2}}2\theta }} $ 

                 $ =\sqrt{2+2\operatorname{Cos}2\theta } $ 

                 $ =\sqrt{2(1+\operatorname{Cos}2\theta )} $ 

                 $ =\sqrt{2.2{{\operatorname{Cos}}^{2}}\theta } $ 

                 $ =2\operatorname{Cos}\theta  $ 

16. Prove that  $ Cot4x\left( Sin5x+Sin3x \right)=Cotx\left( Sin5xSin3x \right) $  

 $ \text{L}\text{.H}\text{.S}\text{.}=Cot4x\left( Sin5x+Sin3x \right) $ 

                $ =\dfrac{\operatorname{Cos}4x}{\operatorname{Sin}4x}\left[ 2\operatorname{Sin}\dfrac{5x+3x}{2}.\operatorname{Cos}\dfrac{5x-3x}{2} \right] $ 

                $ =\dfrac{\operatorname{Cos}4x}{\operatorname{Sin}4x}2\operatorname{Sin}4x.\operatorname{Cos}x $ 

                $ =2\operatorname{Cos}4x.\operatorname{Cos}x $ 

Then, we move on to the right-hand side and using the trigonometric addition identity for the sine function, we obtain

   $ \text{R}\text{.H}\text{.S}\text{.}=Cotx\left( Sin5xSin3x \right) $ 

                   $ =\dfrac{\operatorname{Cos}x}{\operatorname{Sin}x}\left[ 2\operatorname{Cos}\dfrac{5x+3x}{2}.\operatorname{Sin}\dfrac{5x-3x}{2} \right] $ 

                   $ =\dfrac{\operatorname{Cos}x}{\operatorname{Sin}x}[2\operatorname{Cos}4x.\operatorname{Sin}x] $ 

                   $ =2\operatorname{Cos}4x.\operatorname{Cos}x $ 

   $ \text{L}\text{.H}\text{.S}=\text{R}\text{.H}\text{.S} $ 

6 Marks Questions

1. Find the general solution of  $ sin2x+sin4x+sin6x=0 $ 

We have that  $ \sin 2x+\sin 4x+\sin 6x=0 $ 

 $ \Rightarrow \left( \sin 2x+\sin 6x \right)+\sin 4x=0 $ 

 $ \Rightarrow \left( 2\sin \left( \dfrac{2x+6x}{2} \right)\cos \left( \dfrac{2x-6x}{2} \right) \right)+\sin 4x=0 $ 

 $ \Rightarrow 2\sin 4x\cos 2x+\sin 4x=0 $ 

 $ \Rightarrow \sin 4x\left( 2\cos 2x+1 \right)=0 $ 

 $ \Rightarrow x=n\pi  $ 

 $ 2\cos 2x+1=0 $ 

 $ \Rightarrow x=n\pi \pm \dfrac{\pi }{3} $ 

2. Find the general solution of  $ \cos \theta \cos 2\theta \cos 3\theta =\dfrac{1}{4} $  

We have that  $ \cos \theta \cos 2\theta \cos 3\theta =\dfrac{1}{4} $ 

 $ \Rightarrow 4\cos \theta \cos 2\theta \cos 3\theta =1 $ 

Using the trigonometric addition identity for the cosine function, we obtain

 $ \Rightarrow 2\left( 2\cos \theta \cos 3\theta  \right)\cos 2\theta -1=0 $ 

 $ \Rightarrow 2\left( \cos 4\theta +\cos 2\theta  \right)\cos 2\theta -1=0 $ 

 $ \Rightarrow 2\left( 2{{\cos }^{2}}2\theta -1+\cos 2\theta  \right)\cos 2\theta -1=0 $ 

 $ \Rightarrow \left( 2{{\cos }^{2}}2\theta -1 \right)\left( 2\cos 2\theta +1 \right)=0 $ 

 $ 2{{\cos }^{2}}2\theta -1=0 $ 

 $ \Rightarrow \cos 4\theta =0 $ 

 $ \Rightarrow 4\theta =\left( 2n+1 \right)\dfrac{\pi }{2} $ 

 $ \Rightarrow \theta =\left( 2n+1 \right)\dfrac{\pi }{8} $ 

 $ 2\cos 2\theta +1=0 $ 

 $ \Rightarrow \cos 2\theta =-\dfrac{1}{2} $ 

 $ \Rightarrow \theta =n\pi \pm \dfrac{\pi }{3} $ 

3. If  $ \operatorname{Sin}\alpha +\operatorname{Sin}\beta =a $  and  $ \operatorname{Cos}\alpha +\operatorname{Cos}\beta =b $  

Show that  $ \operatorname{Cos}(\alpha +\beta )=\dfrac{{{b}^{2}}-{{a}^{2}}}{{{b}^{2}}+{{a}^{2}}} $  

Squaring both the equations and adding them together,

 $ {{b}^{2}}+{{a}^{2}}={{(\operatorname{Cos}\alpha +\operatorname{Cos}\beta )}^{2}}+{{(\operatorname{Sin}\alpha +\operatorname{Sin}\beta )}^{2}} $ 

                 $ ={{\operatorname{Cos}}^{2}}\alpha +{{\operatorname{Cos}}^{2}}\beta +2\operatorname{Cos}\alpha .\operatorname{Cos}\beta +{{\operatorname{Sin}}^{2}}\alpha +{{\operatorname{Sin}}^{2}}\beta +2\operatorname{Sin}\alpha .\operatorname{Sin}\beta  $ 

                 $ =1+1+2(\operatorname{Cos}\alpha .\operatorname{Cos}\beta +\operatorname{Sin}\alpha .\operatorname{Sin}\beta ) $ 

            $ =2+2\operatorname{Cos}(\alpha -\beta ) $        $ (1) $ 

 $ {{b}^{2}}-{{a}^{2}}={{(\operatorname{Cos}\alpha +\operatorname{Cos}\beta )}^{2}}-{{(\operatorname{Sin}\alpha +\operatorname{Sin}\beta )}^{2}} $ 

                 $ =({{\operatorname{Cos}}^{2}}\alpha -{{\operatorname{Sin}}^{2}}\beta )+({{\operatorname{Cos}}^{2}}\beta -{{\operatorname{Sin}}^{2}}\alpha )+2\operatorname{Cos}(\alpha +\beta ) $ 

                 $ =\operatorname{Cos}(\alpha +\beta )\operatorname{Cos}(\alpha -\beta )+\operatorname{Cos}(\beta +\alpha )\operatorname{Cos}(\alpha -\beta )+2\operatorname{Cos}(\alpha +\beta ) $ 

                 $ =2\operatorname{Cos}(\alpha +\beta ).\operatorname{Cos}(\alpha -\beta )+2\operatorname{Cos}(\alpha +\beta ) $ 

                 $ =\operatorname{Cos}(\alpha +\beta )[2\operatorname{Cos}(\alpha -\beta )+2] $ 

              $ =\operatorname{Cos}(\alpha +\beta ).({{b}^{2}}+{{a}^{2}}) $       from  $ (1) $ 

Dividing equation  $ \left( 1 \right) $  with  $ {{b}^{2}}+{{a}^{2}} $ , we get

 $ \dfrac{{{b}^{2}}-{{a}^{2}}}{{{b}^{2}}+{{a}^{2}}}=\operatorname{Cos}(\alpha +\beta ) $ 

4. Prove  $ Cos\alpha +Cos\beta +Cos\gamma +Cos\left( \alpha +\beta +\gamma  \right)=4\operatorname{Cos}\left( \dfrac{\alpha +\beta }{2} \right).\operatorname{Cos}\left( \dfrac{\beta +\gamma }{2} \right).\operatorname{Cos}\left( \dfrac{\gamma +\alpha }{2} \right) $ 

Starting with the left-hand side and using the trigonometric addition identities for the cosine function, we obtain

 $ \text{L}\text{.H}\text{.S}\text{.}=Cos\alpha +Cos\beta +Cos\gamma +Cos\left( \alpha +\beta +\gamma  \right) $ 

                 $ =2\operatorname{Cos}\left( \dfrac{\alpha +\beta }{2} \right).\operatorname{Cos}\left( \dfrac{\alpha -\beta }{2} \right)+2\operatorname{Cos}\left( \dfrac{\alpha +\beta +\gamma +\gamma }{2} \right).\operatorname{Cos}\left( \dfrac{\alpha +\beta +\gamma -\gamma }{2} \right) $ 

                $ =2\operatorname{Cos}\left( \dfrac{\alpha +\beta }{2} \right).\operatorname{Cos}\left( \dfrac{\alpha -\beta }{2} \right)+2\operatorname{Cos}\left( \dfrac{\alpha +\beta }{2} \right).\operatorname{Cos}\left( \dfrac{\alpha +\beta +2\gamma }{2} \right) $ 

                $ =2\operatorname{Cos}\left( \dfrac{\alpha +\beta }{2} \right)\left[ \operatorname{Cos}\left( \dfrac{\alpha -\beta }{2} \right)+\operatorname{Cos}\left( \dfrac{\alpha +\beta +2\gamma }{2} \right) \right] $ 

            $ =2\operatorname{Cos}\left( \dfrac{\alpha +\beta }{2} \right)\left[ 2\operatorname{Cos}\left( \dfrac{\dfrac{\alpha -\beta }{2}+\dfrac{\alpha +\beta +2\gamma }{2}}{2} \right).\operatorname{Cos}\left( \dfrac{\dfrac{\alpha +\beta +2\gamma }{2}-\dfrac{\alpha -\beta }{2}}{2} \right) \right] $ 

                $ =2\operatorname{Cos}\left( \dfrac{\alpha +\beta }{2} \right)\left[ 2\operatorname{Cos}\left( \dfrac{\alpha +\gamma }{2} \right).\operatorname{Cos}\left( \dfrac{\beta +\gamma }{2} \right) \right] $ 

               $ =4\operatorname{Cos}\left( \dfrac{\alpha +\beta }{2} \right).\operatorname{Cos}\left( \dfrac{\beta +\gamma }{2} \right).\operatorname{Cos}\left( \dfrac{\gamma +\alpha }{2} \right) $ 

5. Prove that  $ \operatorname{Sin}3x+\operatorname{Sin}2x-\operatorname{Sin}2x=4\operatorname{Sin}x.\operatorname{Cos}\dfrac{x}{2}.\operatorname{Cos}\dfrac{3x}{2} $  

 $ \text{L}\text{.H}\text{.S}\text{.}=\sin 3x+\sin x-\sin 2x $ 

                 $ =2\cos \left( \dfrac{3x+x}{2} \right).\operatorname{Sin}\left( \dfrac{3x+x}{2} \right)+\operatorname{Sin}2x $ 

                 $ =2\cos 2x.\sin x+\sin 2x $ 

                 $ =2\cos 2x.\sin x+2\sin x\cos x $ 

                 $ =2\sin x[\cos 2x+\cos x] $ 

                 $ =2\sin x\left[ 2\cos x\dfrac{3x}{2}.\cos \dfrac{x}{2} \right] $ 

                 $ =4\sin x\cos x\dfrac{3x}{2}\cos \dfrac{x}{2} $ 

6. Prove that  $ 2\cos \dfrac{\pi }{13}.\cos \dfrac{9\pi }{13}+\cos \dfrac{3\pi }{13}+\cos \dfrac{5\pi }{13}=0 $  

Starting with the left-hand side using the trigonometric addition identities for the cosine and sine function, we obtain

 $ \text{L}\text{.H}\text{.S}\text{.}=2\cos \dfrac{\pi }{13}.\cos \dfrac{9\pi }{13}+\cos \dfrac{3\pi }{13}+\cos \dfrac{5\pi }{13} $ 

                 $ =\cos \left( \dfrac{\pi }{13}+\dfrac{9\pi }{13} \right)+\cos \left( \dfrac{\pi }{13}-\dfrac{9\pi }{13} \right)+\cos \dfrac{3\pi }{13}+\cos \dfrac{5\pi }{13} $ 

                 $ =\cos \dfrac{10\pi }{13}+\cos \dfrac{18\pi }{13}+\cos \dfrac{3\pi }{13}+\cos \dfrac{5\pi }{13} $ 

                 $ =\cos \left( \pi -\dfrac{3\pi }{13} \right)+\cos \left( \pi -\dfrac{5\pi }{13} \right)+\cos \dfrac{3\pi }{13}+\cos \dfrac{5\pi }{13} $ 

                 $ =-\cos \dfrac{3\pi }{13}-\cos \dfrac{5\pi }{13}+\dfrac{3\pi }{13}+\cos \dfrac{5\pi }{13} $ 

7.  Find the value of  $ \tan (\alpha +\beta ) $ given that  $ \cot \alpha =\dfrac{1}{2},\alpha \in \left( \pi ,\dfrac{3\pi }{2} \right) $  and  $ \operatorname{Sec}\beta =-\dfrac{5}{3},\beta \in \left( \dfrac{\pi }{2},\pi  \right) $  

We know that,

 $ \tan \left( \alpha +\beta  \right)=\dfrac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta } $ 

 $ Cot\alpha =\dfrac{1}{2} $ 

 $ \tan \alpha =2 $ 

Now, let us find  $ \tan \beta  $ 

 $ 1+{{\tan }^{2}}\beta ={{\operatorname{Sec}}^{2}}\beta  $ 

 $ 1+{{\tan }^{2}}\beta ={{\left( \dfrac{-5}{3} \right)}^{2}}\left[ \because \operatorname{Sec}\beta =\dfrac{-5}{3} \right] $ 

 $ \tan \beta =\pm \dfrac{4}{3} $ 

 $ \tan \beta =-\dfrac{4}{3}\left[ \because \beta \in \left( \dfrac{\dfrac{\pi }{2}}{x} \right) \right] $ 

Therefore, we have that

 $ \tan \left( \alpha +\beta  \right)=\dfrac{2-\dfrac{4}{3}}{1-2\left( \dfrac{-4}{3} \right)} $ 

                     $ =\dfrac{2}{11} $ 

Prove that  $ \dfrac{\operatorname{Sec}8A-1}{\operatorname{Sec}4A-1}=\dfrac{\tan 8A}{\tan 2A} $  

Starting with the left-hand side and using the trigonometric elementary identities of the cosine function and sine function, we obtain

 $ \text{L}\text{.H}\text{.S}\text{.}=\dfrac{\sec 8A-1}{\sec 4A-1} $ 

                $ =\dfrac{\dfrac{1}{\operatorname{Cos}8A}-1}{\dfrac{1}{\operatorname{Cos}4A}-1} $ 

                $ =\dfrac{1-\operatorname{Cos}8A}{1-\operatorname{Cos}4A}\times \dfrac{\operatorname{Cos}4A}{\operatorname{Cos}8A} $ 

                $ =\dfrac{2{{\operatorname{Sin}}^{2}}4A}{2{{\operatorname{Sin}}^{2}}2A}.\dfrac{\operatorname{Cos}4A}{\operatorname{Cos}8A} $ 

                $ =\dfrac{(2\operatorname{Sin}4A.\operatorname{Cos}4A).\operatorname{Sin}4A}{2{{\operatorname{Sin}}^{2}}2A.\operatorname{Cos}8A} $ 

                $ =\dfrac{\operatorname{Sin}8A(2\operatorname{Sin}2A.\operatorname{Cos}2A)}{2{{\operatorname{Sin}}^{2}}2A.\operatorname{Cos}8A} $ 

                $ =\dfrac{\operatorname{Sin}8A\operatorname{Cos}2A}{\operatorname{Sin}2A.\operatorname{Cos}2A} $ 

                $ =\dfrac{\tan 8A}{\tan 2A} $ 

Prove that  $ {{\operatorname{Cos}}^{2}}x+{{\operatorname{Cos}}^{2}}\left( x+\dfrac{\pi }{3} \right)+{{\operatorname{Cos}}^{2}}\left( x-\dfrac{\pi }{3} \right)=\dfrac{3}{2} $  

Starting with the left-hand side and using trigonometric addition identities of the cosine function, we obtain

 $ \text{L}\text{.H}\text{.S}\text{.}=\dfrac{1+\operatorname{Cos}2x}{2}+\dfrac{1+\operatorname{Cos}\left( 2x+\dfrac{2\pi }{3} \right)}{2}+\dfrac{1+\operatorname{Cos}\left( 2x-\dfrac{2\pi }{3} \right)}{2} $ 

                $ =\dfrac{1}{2}\left[ 1+1+1+\operatorname{Cos}2x+\operatorname{Cos}\left( 2x+\dfrac{2\pi }{3} \right)+\operatorname{Cos}\left( 2x-\dfrac{2\pi }{3} \right) \right] $ 

                $ =\dfrac{1}{2}\left[ 3+\operatorname{Cos}2x+\operatorname{Cos}\left( 2x+\dfrac{2\pi }{3} \right)+\operatorname{Cos}\left( 2x-\dfrac{2\pi }{3} \right) \right] $ 

                $ =\dfrac{1}{2}\left[ 3+\operatorname{Cos}2x+2\operatorname{Cos}\left( \dfrac{2x+\dfrac{2\pi }{3}+2x-\dfrac{2\pi }{3}}{2} \right).\operatorname{Cos}\left( \dfrac{2x+\dfrac{2\pi }{3}-2x+\dfrac{2\pi }{3}}{2} \right) \right] $ 

                $ =\dfrac{1}{2}\left[ 3+\operatorname{Cos}2x+2\operatorname{Cos}2x.\operatorname{Cos}\dfrac{4\pi }{6} \right] $ 

                $ =\dfrac{1}{2}\left[ 3+\operatorname{Cos}2x+2\operatorname{Cos}2x.\operatorname{Cos}\dfrac{2\pi }{3} \right] $ 

            $ =\dfrac{1}{2}\left[ 3+\operatorname{Cos}2x+2\operatorname{Cos}2x.\operatorname{Cos}\left( \pi -\dfrac{\pi }{3} \right) \right] $ 

                $ =\dfrac{1}{2}\left[ 3+\operatorname{Cos}2x+2\operatorname{Cos}2x.\left( \dfrac{-1}{2} \right) \right] $ 

                $ =\dfrac{3}{2} $ 

 Prove that  $ \operatorname{Cos}2x.\operatorname{Cos}\dfrac{x}{2}-\operatorname{Cos}3x.\operatorname{Cos}\dfrac{9x}{2}=\operatorname{Sin}5x\operatorname{Sin}\dfrac{5x}{2} $  

Starting with the left-hand side and using trigonometric addition identities for the cosine function, we obtain

 $ \text{L}\text{.H}\text{.S}\text{.}=\dfrac{1}{2}\left[ 2\operatorname{Cos}2x.\operatorname{Cos}\dfrac{x}{2}-2\operatorname{Cos}3x.\operatorname{Cos}\dfrac{9x}{2} \right] $ 

                $ =\dfrac{1}{2}\left[ \operatorname{Cos}\left( 2x+\dfrac{x}{2} \right)+\operatorname{Cos}\left( 2x-\dfrac{x}{2} \right)-\operatorname{Cos}\left( \dfrac{9x}{2}+3x \right)-\operatorname{Cos}\left( \dfrac{9x}{2}-3x \right) \right] $ 

                $ =\dfrac{1}{2}\left[ \operatorname{Cos}\dfrac{5x}{2}+\operatorname{Cos}\dfrac{3x}{2}-\operatorname{Cos}\dfrac{15x}{2}-\operatorname{Cos}\dfrac{3x}{2} \right] $ 

               $ =\dfrac{1}{2}\left[ \operatorname{Cos}\dfrac{5x}{2}-\operatorname{Cos}\dfrac{15x}{2} \right] $ 

              $ =\dfrac{1}{2}\left[ -2\operatorname{Sin}\left( \dfrac{\dfrac{5x}{2}+\dfrac{15x}{2}}{2} \right).\operatorname{Sin}\left( \dfrac{\dfrac{5x}{2}-\dfrac{15x}{2}}{2} \right) \right] $ 

              $ =-\operatorname{Sin}5x.\operatorname{Sin}\left( \dfrac{-5x}{2} \right) $ 

              $ =\operatorname{Sin}5x.\operatorname{Sin}\dfrac{5x}{2} $ 

              $ =\text{R}\text{.H}\text{.S}\text{.} $ 

 Prove that  $ \operatorname{Cos}20{}^\circ .\operatorname{Cos}40{}^\circ .\operatorname{Cos}60{}^\circ .\operatorname{Cos}80{}^\circ =\dfrac{1}{16} $  

Starting with the left-hand side and using the trigonometric addition identities of the cosine function, we obtain

 $ \text{L}\text{.H}\text{.S}\text{.}=\operatorname{Cos}{{20}^{{}^\circ }}.\operatorname{Cos}{{40}^{{}^\circ }}.\operatorname{Cos}{{60}^{{}^\circ }}.\operatorname{Cos}{{80}^{{}^\circ }} $ 

                 $ =\operatorname{Cos}{{60}^{{}^\circ }}.\operatorname{Cos}{{20}^{{}^\circ }}.\operatorname{Cos}{{40}^{{}^\circ }}.\operatorname{Cos}{{80}^{{}^\circ }} $ 

                 $ =\dfrac{1}{2}.\dfrac{1}{2}\operatorname{Cos}{{40}^{{}^\circ }}\left( 2\operatorname{Cos}{{20}^{{}^\circ }}.\operatorname{Cos}{{80}^{{}^\circ }} \right) $ 

                 $ =\dfrac{1}{4}\operatorname{Cos}{{40}^{{}^\circ }}\left[ \operatorname{Cos}(80+20)+\operatorname{Cos}(80-20) \right] $ 

                 $ =\dfrac{1}{4}\operatorname{Cos}{{40}^{{}^\circ }}\left[ \operatorname{Cos}{{100}^{{}^\circ }}+\operatorname{Cos}{{60}^{{}^\circ }} \right] $ 

                $ =\dfrac{1}{4}\operatorname{Cos}{{40}^{{}^\circ }}\left[ \operatorname{Cos}{{100}^{{}^\circ }}+\dfrac{1}{2} \right] $ 

                $ =\dfrac{1}{8}(2\operatorname{Cos}{{100}^{{}^\circ }}\operatorname{Cos}{{40}^{{}^\circ }})+\dfrac{1}{8}\operatorname{Cos}{{40}^{{}^\circ }} $ 

           $ =\dfrac{1}{8}\left[ \operatorname{Cos}{{(100+40)}^{{}^\circ }}+\operatorname{Cos}{{(100-40)}^{{}^\circ }} \right]+\dfrac{1}{8}\operatorname{Cos}{{40}^{{}^\circ }} $ 

               $ =\dfrac{1}{8}\left[ \operatorname{Cos}{{140}^{{}^\circ }}+\operatorname{Cos}{{60}^{{}^\circ }} \right]+\dfrac{1}{8}\operatorname{Cos}{{40}^{{}^\circ }} $ 

               $ =\dfrac{1}{8}\left[ \operatorname{Cos}{{140}^{{}^\circ }}+\dfrac{1}{2} \right]+\dfrac{1}{8}\operatorname{Cos}{{40}^{{}^\circ }} $ 

              $ =\dfrac{1}{8}\operatorname{Cos}{{(180-40)}^{{}^\circ }}+\dfrac{1}{16}+\dfrac{1}{8}\operatorname{Cos}{{40}^{{}^\circ }} $ 

              $ =-\dfrac{1}{8}\operatorname{Cos}{{40}^{{}^\circ }}+\dfrac{1}{16}+\dfrac{1}{8}\operatorname{Cos}{{40}^{{}^\circ }} $ 

              $ =\dfrac{1}{16} $ 

             $ =\text{R}\text{.H}\text{.S}\text{.} $ 

 If  $ \tan x=\dfrac{3}{4},\pi <x<\dfrac{3\pi }{2}, $ Find the value of  $ \operatorname{Sin}\dfrac{x}{2},\operatorname{Cos}\dfrac{x}{2} $  and  $ \tan \dfrac{x}{2} $  

 $ \pi <x<\dfrac{3\pi }{2} $  implying that  $ x $  is in the third quadrant 

 $ \Rightarrow \dfrac{\pi }{2}<\dfrac{x}{2}<\dfrac{3\pi }{2} $ 

Therefore, we have that  $ \operatorname{Sin}\dfrac{x}{2} $  is positive and  $ \operatorname{Cos}\dfrac{x}{2} $  is negative.

Let us find for  $ \tan \dfrac{x}{2} $ 

 $ 1+{{\tan }^{2}}x={{\operatorname{Sec}}^{2}}x\dfrac{5}{4} $ 

 $ 1+{{\left( \dfrac{3}{4} \right)}^{2}}={{\operatorname{Sec}}^{2}}x $ 

    $ {{\operatorname{Sec}}^{2}}x=\pm \dfrac{25}{16} $ 

      $ \operatorname{Cos}x=\pm \dfrac{4}{5} $ 

      $ \operatorname{Cos}x=-\dfrac{4}{5}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ \because \pi <x<\dfrac{3\pi }{2} \right] $ 

Let us find the required values 

 $ \operatorname{Sin}\dfrac{x}{2}=\sqrt{\dfrac{1-\operatorname{Cos}x}{2}} $ 

              $ =\sqrt{\dfrac{1+\dfrac{4}{5}}{2}} $ 

              $ =\sqrt{\dfrac{9}{10}} $ 

              $ =\dfrac{3}{\sqrt{10}} $ 

  $ \operatorname{Cos}\dfrac{x}{2}=-\sqrt{\dfrac{1-\operatorname{Cos}x}{2}} $ 

                  $ =-\sqrt{\dfrac{1-\dfrac{4}{5}}{2}} $ 

                  $ =-\sqrt{\dfrac{1}{10}} $ 

                 $ =\dfrac{-1}{\sqrt{10}} $ 

  $ \tan \dfrac{x}{2}=\dfrac{\dfrac{3}{\sqrt{10}}}{\dfrac{-1}{\sqrt{10}}} $ 

                $ =-3 $ 

Class 11 Maths Chapter 3 Important Questions- What are Trigonometric Functions? 

In simple language, the trigonometric functions are the functions of an angle of triangles. It defines the relationship between sides and angles of a triangle is given on basis of these functions. The trigonometric functions consist of the sine, cosine, sectant, cosecant, tangent, and cotangent. It is also known as circular functions. There are several trigonometric formulas and identities that help to define the relationship between the angles and also the functions. At the end of chapter 3, students will find trigonometric functions class 11 extra questions. 

Tips to Score Marks in Trigonometric Functions?

The trigonometric function is one of the important chapters of class 11 maths. The concept of trigonometry was mainly developed to solve geometric problems that incorporate triangles. By practising the important questions of maths class 11 trigonometry functions, students can easily score high marks in the examinations. When students prepare these important questions from Vedantu, they can also learn several tricks and shortcuts to solve the questions fast. Besides, students need to focus on the formulas of trigonometric functions that are crucial to solve the sums. Students should not skip this chapter at any cost as there are many significant areas like designing electronic circuits, finding the heights of tides, etc. To get a deeper insight into class 11 maths ch 3 important questions, students should practice the resources available from Vedantu. 

Discuss the Trigonometric Tables and Formulas?

In class 11 trigonometric functions, important questions about trigonometric tables and formulas constitute a vital part of the chapter. Let’s discuss both these concepts in detail below. 

The Formula for Function of Trigonometric Ratios

Trigonometric Table

Important Questions for Class 11 Maths Chapter 3 Based on Exercise

Q. An engine produces 360 revolutions in one minute. Through how many radians will it turn in one second?

The total number of revolutions made by an engine in one minute = 360

1 minute = 60 seconds

Therefore, number of revolutions in 1 second = 360/60 = 6

Angle formed in 1 revolution = 360°

Angles formed in 6 revolutions = 6 × 360°

Radian measure of the angle in a total of six revolutions = 6 × 360 × π/180

= 6 × 2 × π

So, the engine turns 12π radians in one second.

Importance of Downloading Class 11 Maths Chapter 3 Important Questions PDF?

By downloading the important questions for class 11 maths chapter 3 students will get exposure to the concept of trigonometric functions in depth. Here are some benefits that students will get when they have the PDF version.

They can prepare important notes for the examination. 

They will get access to trigonometry functions class 11 extra questions. 

They can use it as a ready resource for reference. 

It will help them understand the question pattern of the examination. 

Practice Questions

Prove that: (sin 5x + sin 3x)/(cos 5x + cos 3x) = tan 4x

Find the value of tan 765° cot 675° + tan 225° cot 405°

Solve the equation: tan² θ + cot² θ = 2

Write the value of 2sin 75° sin 15°

Show that: tan 4A = (cos8Acos5A - cos12Acos9A) / (sin8Acos5A + cos12Asin9A)

Find the general solution of the following equation: tan2θ +(1 – √3) tan θ – √3 = 0

Prove that: 3sinπ/6secπ/3 - 4sin5π/6cotπ/4 = 1

Find the value: cos 4 π /8 + cos 4 3 π /8 + cos 4 5 π /8 + cos 4 7π/8

Show that: tan 15° + cot 15° = 4

Find the most general value of θ satisfying the equation tan θ = -1 and cos θ = 1/√2

Advantages of Opting Vedantu for Important Questions of Class 11 Maths: Chapter 3 Trigonometric

Vedantu offers several benefits to students using their platform for the "Important Questions for CBSE Class 11 Maths Chapter 3 - Trigonometric Functions (2024-25)":

Comprehensive Coverage: Vedantu's important questions are curated to cover a wide spectrum of topics within the chapter, ensuring a comprehensive understanding of trigonometric functions.

Strategic Exam Preparation : These questions are strategically selected to align with the CBSE curriculum and examination patterns, preparing students effectively for their exams.

Conceptual Clarity: Vedantu's platform emphasizes conceptual clarity by providing in-depth explanations and solutions for each question, helping students grasp the underlying principles.

Variety of Problem Types : The diverse range of questions offered by Vedantu challenges students to apply trigonometric concepts in various problem-solving scenarios, enhancing their problem-solving skills.

Self-Assessment and Practice: Students can use these questions for self-assessment and regular practice, enabling them to gauge their progress and identify areas that need improvement.

Flexibility and Convenience: Vedantu's platform allows students to access these questions anytime, anywhere, providing flexibility in their study routine.

Personalized Learning: Vedantu's adaptive learning approach tailors the learning experience to each student's pace and needs, ensuring effective comprehension and retention.

Interactive Sessions: Vedantu offers interactive live sessions where students can clarify doubts and seek guidance from experienced educators, ensuring a holistic learning experience.

Peer Learning: Students can engage in discussions and peer learning through Vedantu's platform, sharing insights and strategies with their peers.

Holistic Support: Beyond just important questions, Vedantu provides additional study material, revision notes, and comprehensive study plans to support students' overall exam preparation.

Prepare Well with Vedantu Important Questions for CBSE Class 11 Maths Chapter Trigonometric Function 2024-25

The compilation of important questions for CBSE Class 11 Maths Chapter 3 - "Trigonometric Functions" (2024-25) offers a strategic tool to enhance students' grasp of fundamental trigonometric concepts. These questions delve into various aspects of trigonometry, aiding in the development of problem-solving skills and conceptual clarity. By addressing a diverse range of problem types, these questions prepare students comprehensively for examinations and instil a deeper understanding of the subject matter. This resource not only facilitates exam preparation but also promotes a robust foundation for more advanced mathematical exploration, ensuring students' competence in applying trigonometric principles in various scenarios.

Important Related Links for CBSE Class 11 

FAQs on Important Questions for CBSE Class 11 Maths Chapter 3 Trigonometric Functions 2024-25

Q1. What is the easiest way to learn Trigonometry of Class 11 Maths?

Ans: The easiest way to learn Trigonometry is by practice and hard work. Students need to learn all the formulas to solve the questions that might be asked in the exams.  Therefore, the first thing they need to do in the chapter of Trigonometry is to memorize all the formulas by heart and revise them repeatedly, so that it stays in their memory. After this, students must practise all the important questions provided by Vedantu. This will give them an idea of how the question might be asked in the exams. The questions asked in the question paper are never straightforward. Thus, students must have a strong understanding of the concepts to do well in the Class 11 Maths exam.

Q2. Is Trigonometry important for JEE?

Ans: Yes. Trigonometry is one of the most important and prominent topics on which questions are set in the JEE main. Trigonometry functions and Trigonometry ratios are some of the important areas that you might need to focus on. Thus, a good hold on the Trigonometry of Class 11 not only helps students to score more than 90 in their Maths exam but also prepares them for other competitive exams that will shape their future career. Therefore, it goes without saying that students need to practice and work hard in the concepts of Chapter 3 of Class 11 Maths. The NCERT exercises must be practised regularly, to get a better hold of the concepts.

These solutions are available on Vedantu's official website( vedantu.com ) and mobile app free of cost.

Q3. How do you solve Chapter 3 Trigonometry of Class 11 Maths?

Ans: To solve all the numerical problems that might be asked based on the chapter on Trigonometry, students must memorize all the formulas. This will help them solve any question that might be asked in the question paper based on the chapter on Trigonometry. In addition to this, they should practice all the exercises from the NCERT Class 11 Maths Chapter 3. This will prepare and groom them to analyze and solve the questions.

Q4. Is Trigonometry of Class 11 Maths hard?

Ans: Chapter 3 Trigonometry of Class 11 Maths may seem challenging for some students as it has a lot of formulae and derivations to memorize. To score more than 90 in the Class 11 Maths exam, students must practice and revise the important concepts of Chapter 3 regularly. They should also figure out their areas of weaknesses from the chapter and work extra hard on these. Practice is the most important factor to crack this chapter.

Q5. Are the Important Questions for Chapter 3 of Class 11 Maths helpful?

Ans: Vedantu provides Important Questions for Chapter 3 of Class 11 Maths. These are extremely helpful for students as they can help form the base for Class 12. These questions are prepared by experts at Vedantu for the benefit of students. There is a high chance that these questions could be asked in the Class 11 Maths exam. Hence, students should make sure to solve them while they are preparing for their Maths exam, to get a hang of the concepts and the question pattern.

CBSE Class 11 Maths Important Questions

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Important Questions Class 11 Maths Chapter 3

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Important Questions Class 11 Mathematics Chapter 3

Important questions for cbse class 11 mathematics chapter 3 – trigonometric functions.

Trigonometric functions Class 11 Important Questions have been prepared for Class 11 students to help them score higher in the exams. The important Questions for Trigonometric Functions are prepared by subject matter experts in accordance with the latest CBSE guidelines. 

The Class 11 Mathematics Trigonometric Functions Important Questions include step-by-step solutions for questions ranging from simple to difficult. Furthermore, students can easily access these study materials from Extramarks to have a ready reference to the chapters and questions whenever they need it. Students can also make notes and mark them for quick revision based on the important questions. They can easily develop command over trigonometric functions as they answer these important questions.

CBSE Class 11 Mathematics Chapter-3 Important Questions

Study Important Questions for Class 11 Mathematics Chapter 3 – Trigonometric Functions

Students can view the set of important questions given below. 

Q1. Prove that  sin5x−2sin3x+sinx / cos5x−cosx =tanx  

A1. Starting with the left-hand side and using the trigonometric difference identities for the sine function, we obtain

 L.H.S.=sin5x+sinx−2sin3x / cos5x−cosx 

=2sin 3 x.cos 2 x−2sin 3 x /    −2sin 3 x.sin 2 x 

=2sin 3 x(cos 2 x−1)/ −2sin 3 x.sin 2 x 

=−(1−cos 2 x)/−sin 2 x 

=2sin 2 x / 2sinx.cosx 

=sinx / cosx 

Q2. Prove that  cos6x=32cos 2 x − 48cos 4 x + 18cos 2 x − 1  

A2. Starting with the left-hand side and using the trigonometric identities for the cosine function, we obtain

=cos2(3x)=2Cos23x−1 

=2(4cos3x−3cosx)2−1 

=2[16cos6x+9cos2x−24cos4x]−1 

=32cos6x+18cos2x−48cos4x−1 

=32cos6x−48cos4x+18cos2x1 

Q3. Prove that  sin(x+y)/sin(x−y)=tanx+tany/tanx−tany  

A3. Starting with the left-hand side and using the trigonometric difference formula for the sine function, we get

=sin(x+y)/sin(x−y) 

=sinx.cosy+cosx.siny/sinx.cosy−cosx.siny 

Dividing numerator and denominator by  cosx.cosy 

=tanx+tany/tanx−tany 

Q4. The minute hand of a watch is 1.5 cm long. How far does its tip move in 40 minutes? 

A4. Analysing the given information, we have, 

 Angle made in 60min=360 ∘ 

 Angle made in 1min=6 ∘  

 Angle made in 40min=6 ∘ ×40 ∘ =240 ∘  

Calculating the arc distance

240×π/180=l/1.5 

Q6. Show that tan 3x. tan 2x. tan x = tan 3x  tan 2x  tan x  

A6. Let us start with  tan3x  and we know  3x=2x+x 

tan3x=tan(2x+x) 

tan3x/1=tan2x+tanx/1−tan2x.tanx 

tan3x(1−tan2x.tanx)=tan2x+tanx 

tan3x−tan3x.tan2x.tanx=tan2x+tanx 

tan3x.tan2x.tanx=tan3x−tan2x−tanx 

Q7. A wheel makes  360  revolutions in  1  minute. How many radians does it turn in  1  second?

Number of revolutions made in 60s=360 

Number of revolutions made in 1s=360/60 

Angle moved in 6 revolutions = 2π × 6 = 12π 

Given below is the complete set of Important Questions for Trigonometric Functions, which can be accessed by clicking the link provided.

Class 11 Mathematics Chapter 3 Important Questions- What are Trigonometric Functions?

In layman’s terms, trigonometric functions are functions of triangle angles. On the basis of these functions, it defines the relationship between the sides and angles of a triangle. The sine, cosine, secant, cosecant, tangent, and cotangent are trigonometric functions. It is also referred to as circular functions. Several trigonometric formulas and identities can be used to define the relationship between angles and functions. 

Tips to Score Marks in Trigonometric Functions

One of the most important chapters in Class 11 Mathematics is the Trigonometric Function. Trigonometry was developed primarily to solve geometric problems involving triangles. Students can easily score high marks in the examinations by practising the important questions of Mathematics Class 11 Trigonometric Functions. When students prepare for these important questions, they can also learn several tricks and shortcuts to answer the questions quickly. Furthermore, students must concentrate on the trigonometric function formulas that are required to solve the sums. Students should not skip this chapter because it covers many important topics, such as designing electronic circuits, calculating tide heights, and so on.

Discuss the Trigonometric Tables and formulas 

The Formula for Function of Trigonometric Ratios

Trigonometric Table

Important Questions for Class 11 Mathematics Chapter 3 Based on Exercise

• In one minute, an engine makes 360 revolutions. How many radians will it turn in a second?

An engine’s total number of revolutions per minute = 360

1 minute equals 60 seconds

As a result, the number of revolutions in one second = 360 / 60 = 6.

360° is the angle formed in one revolution.

Angles are formed in six revolutions, which equals six 360°.

The angle in radians measured over six revolutions = 6 360 / 180

= 6 × 2 × π

As a result, the engine rotates 12 radians in one second. 

Importance of Downloading Class 11 Mathematics Chapter 3 Important Questions

Students will gain a thorough understanding of Trigonometric Functions by accessing the Important Questions for Class 11 Mathematics Chapter 3 . Here are some advantages:

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Q.1 If A + B = 225°, then tan A + tan B + tan A × tan B is equal to: 1 0 1 3 Marks :1

Ans tan (A + B) = tan (225) = tan (180+45) = tan 45 = 1

Now, tan(A+B) = (tan A+tan B)/(1 tan A × tan B)

(tan A + tan B)/(1? tan A × tan B) = 1

tan A + tan B + tan A × tan B = 1

Q.2 if ? is an acute angle and sin (?/2) = x-12xthen tan ? is Marks :1

Ans tan?=sin?cos?= 2sin?(?/2)?cos?(?/2)1?sin2(?/2)

tan?=2x?12x?1?x?12×1?2x?12x = x2?1

Q.3 Which is greater ? sin1or sin1? Justify your answer. Marks :4

Ans First, we shall convert 1 into degree µ =180°

1=180 =180227 =180—722 =90—711 =63011 1=57.27

sin1=sin 57.27bHence, sin1 is greater than sin1.

Q.4 If three angles A, B, C, are in A.P. Prove that: cotB=sinA-sinCcosC-cosA Marks :3

Ans R.H.S=sinA-sinCcosC-cosA

=2sinA-C2cosA+C22sinA+C2sinA-C2

=cotA+C2=cotB=L.H.S µA,B,C are in A.P  2B=A+C

Q.5 Show that: 2+2+2+2cos8?=2cos? Marks :4

Ans LH.S=2+2+2(1+cos 8¸) =22+2—2cos2 4 µ 1+cos 8 =2cos2 4 =2+2+4cos2 4 =2+2+2cos 4

=2+21+cos4 =2+2-2cos22µ1+cos4 =2cos22=2+4cos22 =2+2cos2 =21+cos2 =2.2cos2 =2cos=R ·H.S

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Faqs (frequently asked questions), 1. what are the fundamental trigonometric functions.

Trigonometric functions are fundamental functions in mathematics that are used to denote the relationship between the angles of a right-angled triangle and the lengths of its sides.

There are six trigonometric ratios in total, with sine, cosine, and tangent serving as the fundamental three. The other three are called cosecant, secant, and cotangent and are described in relation to previously discussed basic functions.

In a given right-angled triangle, the following values of these ratios are calculated for a specific angle θ in terms of its sides:

sin θ = opposite side / hypotenuse

cos θ = adjacent side / hypotenuse

tan θ = sin θ / cos θ = opposite site / adjacent side

cosec θ = 1 / sin θ

sec θ = 1 / cos θ

cot θ = cos θ / sin θ = 1 / tan θ

2. What is the relationship between Radians and Degrees?

A radian (abbreviated rad or c) is a standard unit of measurement for angles based on the relationship between the length of an arc and the radius of a circle.

rad = arc length / circle radius

Many times, you will be required to convert given angle values  from radians to degrees. The relationship between these two units can be deduced as follows.

The circumference C of a full circle with radius r and angle 360o is equal to the arc length.

Therefore, arc length = C = 2πr.

From the first relation, 360o in radians will be

rad = 2πr / r = 2π

Therefore, 2π rad = 360o,

that is, 1 rad = 180o / π.

This brings us to the final relationship, that is, degree = radian x 180 / π.

3. What are the practical applications of trigonometry?

Trigonometry has a wide range of applications in marine biology, navigation, aviation, and other fields because it involves the relationship between sides and angles, finding the height, and calculating the distance. It is also used to solve complex mathematical problems, such as those in calculus and algebra.

4. What are the angles of trigonometry?

You will come across several new terms that are used in trigonometry questions as you study Chapter 3 Trigonometry of Class 11 Mathematics. The angles of trigonometry are the most important of these. These are essentially functions that aid in the relationship of triangle sides and angles. The sine, cosine, and tangent are abbreviated as sin, cos, and tan. Secant, cosecant, and cotangent are the inverse functions or angles, abbreviated as sec, cosec, and cot.

5. Name the Six Trigonometric Functions.

The six trigonometric functions are sine, cosine, tan, cosec, sec, and cot.

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NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions PDF Download

NCERT Solutions are essential for all students who want to score well in their exams. Our subject matter experts have developed these solutions to help the students score good marks in their examinations. These solutions are well-explained and include each and every topic. Another essential benefit of NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions is that it is created as per the latest pattern of CBSE (Central Board of Secondary Education). These solutions also ensure that a student gets a clear understanding of each and every concept. 

In the Class 11 Maths Chapter 3 Trigonometric Functions Miscellaneous Solutions, each and every type of question is included. From easy to hard, the students will get an idea of all the types of questions which can appear in the exam. Also, apart from this, the students will also get to know about the most common questions which have a high chance of appearing in the exam. 

It can be assumed that a student will score good marks in their exams if they will study the Maths Class 11 Chapter 3 Trigonometric Functions PDF NCERT Solution on a regular basis. Our subject matter experts at Selfstudys have invested a good amount of time in creating them so that the students do not find any difficulty studying from these solutions. As the experts have been in the education line for quite some time, they are aware of the learning potential of every student, which is why they have created these solutions in a simple way so that no student faces any difficulty studying from them. Students can also use these solutions at the time of revision. 

The NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions can be easily downloaded from the website of Selfstudys i.e. Selfstudys.com. 

NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions PDF 

The NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions PDF can be easily downloaded from the website of Selfstudys. As they are in PDF format, it becomes easier for the students to access them. They are also mobile-friendly. 

How to Download the NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions? 

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• Following that, you will be redirected to a page where you need to select the class and the subject for which you want to download the NCERT Solutions. 
• In this case, you need to select class 10 and the subject Maths to download the NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions. 

What are the Features of the Maths Class 11 Chapter 3 Trigonometric Functions PDF NCERT Solution? 

There are many features of the NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions which can be very beneficial for all the students. The most important features include: 

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The NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions has several benefits to offer to the students. Some of the most important of them are 

• Clear Understanding of the Topic: The Class 11 Maths Chapter 3 Trigonometric Functions Miscellaneous Solutions provides in-depth knowledge of each and every topic which is present in the syllabus. Each chapter is explained in detail in a step-by-step process. To make learning easier and more interesting for all the students, diagrams and examples are also given. 
• Authentic Solutions: The Maths Class 11 Chapter 3 Trigonometric Functions PDF NCERT Solution are completely authentic and can be trusted 100%. The highly qualified subject matter experts at Selfstudys have created these solutions in an authentic way which makes it simple for all the students to learn all the concepts. 
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• Conceptual Clarity: The NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions focuses on building a strong foundation for all the students. Students can get a deep understanding of the fundamental principles of maths formulas, which serve as a basis for further mathematical topics. 
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What are the tips to study from the NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions? 

There are various tips that can help the students to study effectively from the NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions: 

• Go through the chapter thoroughly: It is very important for all the students to first go through the chapter as it will help them to understand the basics. If their basics will get strong, they will automatically be able to grasp the other concepts in a better way. 
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• Do not get distracted: All the students should stay focused to concentrate better and should not get distracted. They should stay away from their mobile phones and other electronic gadgets, for example, TV, etc. to avoid all the distractions while studying Maths Class 11 Chapter 3 Trigonometric Functions PDF NCERT Solution. 
• Practice Time Management: While doing exam preparation from the NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions, it is advisable for all the students to practice time management by setting a time duration for solving problems. This will help you develop efficiency and improve your speed in solving problems accurately. 

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There are various ways to master NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions. Some of the most important of them are: 

• Study the NCERT Solutions: The students can start by studying the Class 11 Maths Chapter 3 Trigonometric Functions Miscellaneous Solutions. It is advisable for them to read the solutions carefully and understand them stepwise manner used to solve each problem. They should use them regularly as it will help them to understand the concepts in a clear manner. 
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• Do group discussions with your peers: It is advisable for all the students to do group discussions with their peers as discussing the concepts, problem-solving techniques, and solutions with others can help you understand them in a better way. Teaching and explaining concepts to others can deepen your understanding of all the concepts. 
• Revise regularly: Regularly revise the concepts and solutions from NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions. Review your notes, summaries, and solved problems. Revisit the solutions and attempt to solve them again without referring to the provided solutions. This revision will help reinforce your understanding and improve retention. 
• Take note of your mistakes and work on them: It is advisable for all the students to take note of their mistakes while solving the Maths Class 11 Chapter 3 Trigonometric Functions PDF NCERT Solution and work on them. Find your weaknesses and identify the specific concepts or steps that caused confusion. This will help you improve your problem-solving skills. 

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To maximize your learning potential while solving the NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions, here are the tips which can help you: 

• Do a thorough reading of the chapter: All the students should read Maths Class 11 Chapter 3 Trigonometric Functions PDF NCERT Solution thoroughly to maximize their learning potential as it will help them to understand the basic concepts which will make it easier for them to grasp all the concepts. 
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• Revise the concepts on a regular basis: It is very important for all the students to Revise all the concepts from the NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions as it helps to recall all the information which they have studied during the preparation for the exam. You can also use different methods to revise the concepts on a regular basis, for example, making flashcards, notes, etc. 

What Are the Methods To Study From NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions? 

Studying for the exam requires focus and concentration. Below we will discuss various methods of how students can study for an Exam with the help of NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions: 

• Making a strong grip of basics: It is very important for all students to create a strong base for basics and learn them for NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions as it helps a student to get a clear understanding of the concepts. This makes it easier for the students to learn them fast. 
• Study the NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions on a regular basis: It is advisable for all the students to study Class 11 Maths Chapter 3 Trigonometric Functions Miscellaneous Solutions on a regular basis as it will help them to do effective preparation for their exams and also increases the chances of the students to score good marks in their exams. 
• Always read the summary of your textbook: It is advisable for all students to always read the summary which is included in the Maths Class 11 Chapter 3 Trigonometric Functions PDF NCERT Solution. The students should read them on a regular basis as it will help them to get a thorough knowledge of all the topics and will also help them to prepare well for the exams. This will increase their chances of scoring good marks in the exams. 
• Identify your strong and weak areas: It is very important for all the students to identify their strong and weak areas as it will help them to understand where they need to put more effort in order to do well in their exams. 
• Do timely revision: All the students should do timely revision from the NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions if they want to do extremely well in their exams as it will ensure that you have created a strong base of all the topics in their minds. 

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Download Case Study Questions for Class 11 Maths

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[PDF] Download Case Study Questions for Class 11 Maths

Here we are providing case study questions for Class 11 Maths. In this article, we are sharing links for Class 11 Maths All Chapters. All case study questions of Class 11 Maths are solved so that students can check their solutions after attempting questions.

Click on the chapter to view.

Class 11 Maths Chapters

Chapter 1 Sets Chapter 2 Relations and Functions Chapter 3 Trigonometric Functions Chapter 4 Principle of Mathematical Induction Chapter 5 Complex Numbers and Quadratic Equations Chapter 6 Linear Inequalities Chapter 7 Permutation and Combinations Chapter 8 Binomial Theorem Chapter 9 Sequences and Series Chapter 10 Straight Lines Chapter 11 Conic Sections Chapter 12 Introduction to Three-Dimensional Geometry Chapter 13 Limits and Derivatives Chapter 14 Mathematical Reasoning Chapter 15 Statistics Chapter 16 Probability

What is meant by Case Study Question?

In the context of CBSE (Central Board of Secondary Education), a case study question is a type of question that requires students to analyze a given scenario or situation and apply their knowledge and skills to solve a problem or answer a question related to the case study.

Case study questions typically involve a real-world situation that requires students to identify the problem or issue, analyze the relevant information, and apply their understanding of the relevant concepts to propose a solution or answer a question. These questions may involve multiple steps and require students to think critically, apply their problem-solving skills, and communicate their reasoning effectively.

Importance of Solving Case Study Questions for Class 11 Maths

Case study questions are an important aspect of mathematics education at the Class 11 level. These questions require students to apply their knowledge and skills to real-world scenarios, helping them develop critical thinking, problem-solving, and analytical skills. Here are some reasons why case study questions are important in Class 11 maths education:

• Real-world application: Case study questions allow students to see how the concepts they are learning in mathematics can be applied in real-life situations. This helps students understand the relevance and importance of mathematics in their daily lives.
• Higher-order thinking: Case study questions require students to think critically, analyze data, and make connections between different concepts. This helps develop higher-order thinking skills, which are essential for success in both academics and real-life situations.
• Collaborative learning: Case study questions often require students to work in groups, which promotes collaborative learning and helps students develop communication and teamwork skills.
• Problem-solving skills: Case study questions require students to apply their knowledge and skills to solve complex problems. This helps develop problem-solving skills, which are essential in many careers and in everyday life.
• Exam preparation: Case study questions are included in exams and tests, so practicing them can help students prepare for these assessments.

Overall, case study questions are an important component of Class 11 mathematics education, as they help students develop critical thinking, problem-solving, and analytical skills, which are essential for success in both academics and real-life situations.

Feature of Case Study Questions on This Website

Here are some features of a Class 11 Maths Case Study Questions Booklet:

Many Case Study Questions: This website contains many case study questions, each with a unique scenario and problem statement.

Different types of problems: The booklet includes different types of problems, such as optimization problems, application problems, and interpretation problems, to test students’ understanding of various mathematical concepts and their ability to apply them to real-world situations.

Multiple-choice questions: Questions contains multiple-choice questions to assess students’ knowledge, understanding, and critical thinking skills.

Focus on problem-solving skills: The questions are designed to test students’ problem-solving skills, requiring them to identify the problem, select appropriate mathematical tools, and analyze and interpret the results.

Emphasis on practical applications: The case studies in the booklet focus on practical applications of mathematical concepts, allowing students to develop an understanding of how mathematics is used in real-life situations.

Comprehensive answer key: The booklet includes a comprehensive answer key that provides detailed explanations and step-by-step solutions for all the questions, helping students to understand the concepts and methods used to solve each problem.

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NCERT Solutions for Class 6, 7, 8, 9, 10, 11 and 12

NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions

September 22, 2019 by phani

Get Free NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.1, Ex 3.2, Ex 3.3, Ex 3.4, and Miscellaneous Exercise PDF in Hindi and English Medium. Trigonometric Functions Class 11 Maths NCERT Solutions are extremely helpful while doing your homework. Trigonometric Functions All Exercises Class 11 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers.

Class 11 Maths Trigonometric Functions NCERT Solutions in English Medium and Hindi Medium

• Trigonometric Functions Class 11 Ex 3.1
• Trigonometric Functions Class 11 Ex 3.2
• Trigonometric Functions Class 11 Ex 3.3
• Trigonometric Functions Class 11 Ex 3.4
• Trigonometric Functions Class 11 Miscellaneous Exercise

त्रिकोणमितीय फलन प्रश्नावली 3.1 का हल हिंदी में

• त्रिकोणमितीय फलन प्रश्नावली 3.2 का हल हिंदी में
• त्रिकोणमितीय फलन प्रश्नावली 3.3 का हल हिंदी में
• त्रिकोणमितीय फलन प्रश्नावली 3.4 का हल हिंदी में
• त्रिकोणमितीय फलन विविध प्रश्नावली का हल हिंदी में
• Trigonometry Formulas
• Trigonometry Functions Class 11 Notes
• NCERT Exemplar Class 11 Maths Chapter 3 Trigonometric Functions
• Trig Cheat Sheet
• JEE Main Trigonometry Previous Year Questions

Free download NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.1, Ex 3.2, Ex 3.3, Ex 3.4, and Miscellaneous Exercise PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, BIE, Intermediate and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2019-20.

NCERT Solutions for Class 11 Maths Chapter 3 Exercise 3.1

Ex 3.1 Class 11 Maths Question 1: Find the radian measures corresponding to the following degree measures: (i) 25° (ii) – 47° 30′ (iii) 240° (iv) 520° Ans:

(i) \(\frac{11}{16}\) We know that: π radian = 180° ∴ \(\frac{11}{16}\) radain = \(\frac{180}{\pi} \times \frac{11}{16}\) × degree

= \(\frac{45 \times 11}{\pi \times 4}\) degree

= \(\frac{45 \times 11 \times 7}{22 \times 4}\) degree

= \(\frac{315}{8}\) degree

= 39 \(\frac{3}{8}\) degree

= 39° + \(\frac{3 \times 60}{8}\) minutes [1° = 60′]

= 39° + 22′ + \(\frac{1}{2}\) minutes

= 39°22’30” [1′ = 60°].

More Resources for CBSE Class 11

NCERT Solutions

• NCERT Solutions Class 11 Maths
• NCERT Solutions Class 11 Physics
• NCERT Solutions Class 11 Chemistry
• NCERT Solutions Class 11 Biology
• NCERT Solutions Class 11 Hindi
• NCERT Solutions Class 11 English
• NCERT Solutions Class 11 Business Studies
• NCERT Solutions Class 11 Accountancy
• NCERT Solutions Class 11 Psychology
• NCERT Solutions Class 11 Entrepreneurship
• NCERT Solutions Class 11 Indian Economic Development
• NCERT Solutions Class 11 Computer Science

Ex 3.1 Class 11 Maths Question 3: A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second? Ans: Number of revolutions made by the wheel in 1 minute = 360 ∴ Number of revolutions made by the wheel in 1 second = \(\frac{360}{6}\) = 6 In one complete revolution, the wheel turns an angle of 2π radian. Hence, in 6 complete revolutions, it will turn an angle of 6 × 2π radian, i.e., 12π radian Thus, in one second, the wheel turns an angle of 12π radian. Ex 3.1 Class 11 Maths Question 4: Find the degree measure of the angle subtended at the centre of a circle of radius 100 cm y an arc of length 22 cm (Use π = \(\frac{22}{7}\)). Ans:

We know that in a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then θ = \(\frac{l}{r}\) Therefore, for r = 100 cm, l = 22 cm, we have

Thus, the required angle is 12°36′.

Ex 3.1 Class 11 Maths Question 5: In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of minor arc of the chord. Ans:

Given, diameter = 40 cm ∴ radius (r) = \(\frac{40}{2}\) = 20 cm and length of chord, AB = 20 cm Thus, ∆OAB is an equilateral triangle. We know that, θ = \(\frac{\text { Arc } A B}{\text { radius }}\) ⇒ Arc AB = θ × r = \(\frac{\pi}{3}\) × 20 . = \(\frac{20}{3}\) π cm.

Ex 3.1 Class 11 Maths Question 6: If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, find the ratio of their radii. Ans:

Let the radii of the two circles be r 1  and r 2 . Let an arc of length l subtend an angle of 60° at the centre of the circle of radius r 1 , while let an arc of length l subtend an angle of 75° at the centre of the circle of radius r 2 . Now, 6o° = \(\frac{\pi}{3}\) radian and 75° = \(\frac{5 \pi}{12}\) radian We know that in a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then θ = \(\frac{l}{r}\) or l = rθ ∴ l = \(\frac{r_{1} \pi}{3}\) and

l = \(\frac{r_{2} 5 \pi}{12}\)

⇒ \(\frac{r_{1} \pi}{3}=\frac{r_{2} 5 \pi}{12}\)

⇒ r = \(\frac{r_{2} 5}{4}\)

\(\frac{r_{1}}{r_{2}}=\frac{5}{4}\) Thus, the ratio of the radii is 5 : 4.

Ex 3.1 Class 11 Maths Question 7: Find the angle in radian through which a pendulum swings if its length is 75 cm and the tip describes an arc of length (i) 10 cm (ii) 15 cm (iii) 21 cm. Ans:

We know that in a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then θ = \(\frac{l}{r}\). It is given that r = 75 cm

(i) Here, l = 10 cm θ = \(\frac{10}{75}\) radian = \(\frac{2}{15}\) radian

(ii) Here, l = 15 cm θ = \(\frac{15}{75}\) radian θ = \(\frac{1}{5}\) radian

(iii) Here, l = 21 cm θ = \(\frac{21}{75}\) radian = \(\frac{7}{75}\) radian.

NCERT Solutions for Class 11 Maths Chapter 3 Exercise 3.2

sin x = \(\frac{3}{5}\)

cosec x = \(\frac{1}{\sin x}=\frac{1}{\left(\frac{3}{5}\right)}=\frac{5}{3}\) sin 2  x + cos 2  x = 1 ⇒ cos 2  x = 1 – sin 2  x ⇒ cos 2  x = 1 – (\(\frac{3}{5}\)) 2

⇒ cos 2  x = 1 – \(\frac{9}{25}\)

⇒ cos 2  x = \(\frac{16}{25}\)

⇒ cos x = ± \(\frac{4}{5}\) Since x lies in the 2nd quadrant, the value of cos x will be negative

⇒ \(\frac{4}{3}=\frac{\sin x}{\frac{-3}{5}}\) ⇒ sin x = \(\left(\frac{4}{3}\right) \times\left(\frac{-3}{5}\right)=-\frac{4}{5}\) ⇒ cosec x = \(\frac{1}{\sin x}=-\frac{5}{4}\).

tan x = – \(\frac{5}{12}\)

cot x = \(\frac{1}{\tan x}=\frac{1}{\left(-\frac{5}{12}\right)}=-\frac{12}{5}\)

Ex 3.2 Class 11 Maths Question 6: Find the value of the trigonometric function sin 765°. Ans: It is known that the values of sin x repeat after an interval of 2π or 360°. ∴ sin 765° = sin (2 × 360° + 45°) = sin 45° = 1 Ex 3.2 Class 11 Maths Question 7: Find the value of the trigonometric function cosec (- 1410°) Ans: It is known that the values of cosec x repeat after an interval of 2π or 360°. ∴ cosec (- 1410°) = cosec (- 1410° + 4 x 360°) = cosec (- 1410° + 1440°) = cosec 30° = 2. Ex 3.2 Class 11 Maths Question 8: Find the value of the trigonometric function tan \(\frac{19 \pi}{3}\). Ans:

It is known that the values of tan x repeat after an interval of π or 180°. ∴ \(\tan \frac{19 \pi}{3}=\tan 6 \frac{1}{3} \pi\)

= \(\tan \left(6 \pi+\frac{\pi}{3}\right)=\tan \frac{\pi}{3}\)

= tan 60° = √3.

Ex 3.2 Class 11 Maths Question 9: Find the value of the trigonometric function sin \(\left(-\frac{11 \pi}{3}\right)\). Ans:

It is known that the values of cot x repeat after an interval of π or 180°.

∴ \(\sin \left(\frac{11 \pi}{3}\right)=\sin \left(-\frac{11 \pi}{3}+2 \times 2 \pi\right)\)

= \(\sin \left(\frac{\pi}{3}\right)=\sin 60^{\circ}=\frac{\sqrt{3}}{2}\)

Ex 3.2 Class 11 Maths Question 10: Find the value of the trigonometric function cot \(\left(-\frac{15 \pi}{4}\right)\). Ans:

It is known that the values of cot x repeat after an interval of ir or 1800. ∴ \(\cot \left(-\frac{15 \pi}{4}\right)=\cot \left(-\frac{15 \pi}{4}+4 \pi\right)=\cot \frac{\pi}{4}\) = 1.

NCERT Solutions for Class 11 Maths Chapter 3 Exercise 3.3

Ex 3.3 Class 11 Maths Question 1:

Prove that: sin 2  \(\frac{\pi}{6}\) + cos 2  \(\frac{\pi}{3}\) – tan 2  \(\frac{\pi}{4}\) = – \(\frac{1}{2}\) Ans:

L.H.S.= sin 2  \(\frac{\pi}{6}\) + cos 2  \(\frac{\pi}{3}\) – tan 2  \(\frac{\pi}{4}\)

= \(\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}\) – (1) 2

= \(\frac{1}{4}+\frac{1}{4}-1=-\frac{1}{2}\)

= R.H.S. Hence proved.

L.H.S = \(2 \sin ^{2} \frac{3 \pi}{4}+2 \cos ^{2} \frac{\pi}{4}+2 \sec ^{2} \frac{\pi}{3}\)

= \(2\left\{\sin \left(\pi-\frac{\pi}{4}\right)\right\}^{2}+2\left(\frac{1}{\sqrt{2}}\right)^{2}+2(2)^{2}\)

= \(2\left\{\sin \frac{\pi}{4}\right\}^{2}+2 \times \frac{1}{2}+8\)

= 2 \(\left(\frac{1}{\sqrt{2}}\right)^{2}\) + 1 + 8

= 1 + 1 + 8 = 10 = R.H.S. Hence proved.

Ex 3.3 Class 11 Maths Question 5: Find the value of: (i) sin 75°, (ii) tan 15° Ans:

(i) sin 75° sin (45° + 30°) = sin 45° cos 30° + cos 45° sin 30° [∵ sin (x + y) = sin x cos y + cos x sin y] = \(\left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3}}{2}\right)+\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{2}\right)\)

= \(\frac{\sqrt{3}}{2 \sqrt{2}}+\frac{1}{2 \sqrt{2}}=\frac{\sqrt{3}+1}{2 \sqrt{2}}\)

(ii) tan 15° = tan (45° – 30°)

L.H.S = \(\frac{\cos (\pi+x) \cos (-x)}{\sin (\pi-x) \cos \left(\frac{\pi}{2}+x\right)}\)

= \(\frac{[-\cos x][\cos x]}{(\sin x)(-\sin x)}=\frac{-\cos ^{2} x}{-\sin ^{2} x}\)

= R.H.S Hence proved.

It is known that cos A – cos B = \(-2 \sin \left(\frac{A+B}{2}\right) \cdot \sin \left(\frac{A-B}{2}\right)\)

∴ L.H.S.= \(=\cos \left(\frac{3 \pi}{4}+x\right)-\cos \left(\frac{3 \pi}{4}-x\right)\)

= \(– 2 \sin \left\{\frac{\left(\frac{3 \pi}{4}+x\right)+\left(\frac{3 \pi}{4}-x\right)}{2}\right\} \cdot \sin \left\{\frac{\left(\frac{3 \pi}{4}+x\right)-\left(\frac{3 \pi}{4}-x\right)}{2}\right\}\)

= – 2 sin (\(\frac{3 \pi}{4}\)) sin x

= – 2 sin (- \(\frac{\pi}{4}\)) sin x

= – √2 sin x = R.H.S. Hence proved.

Ex 3.3 Class 11 Maths Question 12: Prove that: sin 2  6x – sin 2  4x = sin 2x sin 10 x Ans:

It is known that sin A + sin B = 2 \(\sin \left(\frac{A-B}{2}\right) \cos \left(\frac{A-B}{2}\right)\)

sin A – sin B = 2 \(\cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)\) L.H.S.= sin 2  6x – sin 2  4x = (sin 6x + sin 4x) (sin 6x – sin 4x) = (2 sin 5x cos x) (2 cos 5x sin x) = (2 sin 5x cos 5x) (2 sin x cos x) = sin 10x sin 2x = R.H.S. Hence proved.

Ex 3.3 Class 11 Maths Question 13: Prove that: cos 2  2x cos 2  6x = sin 4x sin 8x Ans:

It is known that cos A + cos B = 2 \(\cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\)

cos A – cos = 2 \(\sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)\)

∴ L.H.S = cos 2  2x – cos 2  6x = (cos 2x + cos 6x) (cos 2x – 6x) = \(\left[2 \cos \left(\frac{2 x+6 x}{2}\right) \cos \left(\frac{2 x-6 x}{2}\right)\right]\left[-2 \sin \left(\frac{2 x+6 x}{2}\right) \sin \frac{(2 x-6 x)}{2}\right]\) ∴ L.H.S.= cos 2  2x – cos 2  6x = (cos 2x + cos 6x) (cos 2x – 6x) = [2 cos 4x cos (-2x)] [- 2 sin 4x sin (- 2x)] = [2 cos 4x cos 2x] [- 2 sin 4x (- sin 2x)] = (2 sin 4x cos 4x) (2 sin 2x cos 2x) = sin 8x sin 4x = R.H.S. Hence proved.

It is known that sin A + sin = 2 \(\sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\)

cos A + cos = 2 \(\cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\)

∴ L.H.S = \(\frac{\sin 5 x+\sin 3 x}{\cos 5 x+\cos 3 x}\)

= \(\frac{2 \sin \left(\frac{5 x+3 x}{2}\right) \cos \left(\frac{5 x-3 x}{2}\right)}{2 \cos \left(\frac{5 x+3 x}{2}\right) \cos \left(\frac{5 x-3 x}{2}\right)}\)

= \(\frac{2 \sin 4 x \cos x}{2 \cos 4 x \cos x}=\frac{\sin 4 x}{\cos 4 x}\)

= tan 4x = R.H.S. Hence proved.

It is known that sin A – sin B = 2 \(\cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)\)

cos 2  A – sin 2  A = cos 2A

∴ L.H.S. = \(=\frac{\sin x-\sin 3 x}{\sin ^{2} x-\cos ^{2} x}\)

= \(\frac{2 \cos \left(\frac{x+3 x}{2}\right) \sin \left(\frac{x-3 x}{2}\right)}{-\cos 2 x}\)

= \(\frac{2 \cos 2 x \sin (-x)}{-\cos 2 x}\)

= – 2 × (- sin x) = 2 sin x

Hence proved.

Class 11 Maths NCERT Solutions Chapter 3 Exercise 3.4

cos 4x = cos 2x cos 4x – cos 2x = 0 – 2 sin \(\left(\frac{4 x+2 x}{2}\right)\) sin \(\left(\frac{4 x-2 x}{2}\right)\) = 0

[∵ cos A – cos B = 2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)\(\)]

sin 3x sin x = 0 sin 3x = 0or sin x = 0 3x = nπ or x = nπ, where n ∈ Z x = \(\frac{n \pi}{3}\) or x = nπ, where n ∈ Z.

Ex 3.4 Class 11 Maths Question 6: Find the general solution of the equation cos 3x + cosx – cos 2x = 0 Ans:

cos 3x + cos x – cos 2x = 0 2 cos \(\left(\frac{3 x+x}{2}\right)\) cos \(\left(\frac{3 x-x}{2}\right)\) – cos 2x = 0

[∵ cos A + cos B = 2 \(\cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\)]

2 cos 2x cos x – cos 2x = 0 cos 2x (2 cos x – 1) = 0 cos 2x = 0 or 2 cos x – 1 = 0 cos 2x = 0 or cos x = \(\frac{1{2}\) ∴ 2x = (2n + 1) \(\frac{\pi}{2}\) or cos x = cos \(\frac{\pi}{3}\), where n ∈ Z x = (2n + 1) \(\frac{\pi}{4}\) or x = 2nπ ± \(\frac{\pi}{3}\) where n ∈ Z.

Ex 3.4 Class 11 Maths Question 7: Find the general solution of the equation sin 2x + cos x = 0 Ans:

sin 2x + cos x = 0 ⇒ 2sin x cos x + cos x = 0 ⇒ cos x (2 sin x + 1) = 0 ⇒ cos x = 0 or 2 sin x + 1 = 0 Now, cos x = 0 ⇒ x = (2n + 1) \(\frac{\pi}{2}\) , where n ∈ Z. or 2 sin x + 1 = 0 ⇒ sin x = – \(\frac{1}{2}\)

= – sin \(\frac{\pi}{6}\)

= sin (π + \(\frac{\pi}{6}\))

= sin \(\frac{7 \pi}{6}\)

x = nπ + (- 1) n  \(\frac{7 \pi}{6}\) where n ∈ Z Therefore, the general solution is (2n + 1) \(\frac{\pi}{2}\) or nπ + (- 1) n  \(\frac{7 \pi}{6}\) where n ∈ Z.

Ex 3.4 Class 11 Maths Question 8: Find the general solution of the equation sec 2  2x = 1 – tan 2x. Ans:

sec 2  2x = 1 – tan 2x 1 + tan 2  2x = 1 – tan 2x tan 2  x + tan 2x = 0 => tan 2x (tan 2x + 1) = 0 => tan 2x = 0 or tan 2x + 1 = 0 Now, tan 2x = 0 => tan 2x = tan 0 2x = nπ + 0, where n ∈ Z x = \(\frac{n \pi}{2}\), where n ∈ Z or tan 2x + 1 = 0 = tan 2x = – 1 = – tan \(\frac{\pi}{4}\)

= tan (π – \(\frac{\pi}{4}\))

= tan \(\frac{3 \pi}{4}\)

2x = nπ + \(\frac{3 \pi}{4}\) where n ∈ Z

x = \(\frac{n \pi}{2}+\frac{3 \pi}{8}\), where n ∈ Z

Therefore, the general solution is \(\frac{n \pi}{2}\) or \(\frac{n \pi}{2}+\frac{3 \pi}{8}\) where n ∈ Z.

Ex 3.4 Class 11 Maths Question 9: Find the general solution of the equation sin x + sin 3x + sin 5x = 0 Ans:

sin x + sin 3x + sin 5x = 0 ⇒ (sin x + sin 5x) + sin 3x = 0

\(\left[2 \sin \left(\frac{x+5 x}{2}\right) \cos \left(\frac{x-5 x}{2}\right)\right]\) + sin 3x = 0

[∵ sin A + sin B = 2 sin \(\sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\)]

2 sin 3x cos (2x) + sin 3x = 0 2 sin 3x cos 2x + sin 3x = 0 sin 3x (2 cos 2x +1) = 0 sin 3x = 0 or 2 cos 2x + 1 = 0 Now sin 3x = 0 ⇒ 3x = nπ, where n ∈ Z i.e., x = \(\frac{n \pi}{3}\) where n ∈ Z or 2 cos 2x + 1 = 0 cos 2x = \(-\frac{1}{2}\)

= – cos \(\frac{\pi}{3}\)

= cos (π – \(\frac{\pi}{3}\))

cos 2x = cos \(\frac{2 \pi}{3}\)

⇒ 2x = 2nπ ± \(\frac{2\pi}{3}\), where n ∈ Z

⇒ x = nπ ± \(\frac{\pi}{3}\), where n ∈ Z

Therefore, the general solution is \(\frac{n \pi}{3}\) or nπ ± \(\frac{\pi}{3}\), where n ∈ Z.

NCERT Solutions for Class 11 Maths Chapter 3 Miscellaneous Exercise

L.H.S. = (sin 3x + sin x) sin x + (cos 3x – cos x) cos x = sin 3x sin x + sin 2  x + cos 3x cos x – cos 2  x = cos 3x cos x + sin 3x sin x – (cos 2  x – sin 2  x) = cos (3x – x) – cos 2x [∵ cos(A – B) = cos A cos B + sin A sin B] = cos 2x – cos 2x = 0 =R.H.S. Hence proved.

Miscellaneous Exercise Class 11 Maths Question 3:

Prove that: (cos x + cos y) 2  + (sin x – sin y) 2  = 4 cos 2  \(\left(\frac{x+y}{2}\right)\)

L.H.S.= (cos x + cos y) 2  + (sin x – sin y) 2 = cos 2  x + cos 2  y + 2 cos x cos y + sin 2  x + sin 2  y – 2 sin x sin y = (cos 2  x + sin 2  x) + (cos 2  y + sin 2  y) + 2 (cos x cos y – sin x sin y) = 1 + 1 + 2 cos (x + y) [∵ cos (A + B) = (cos A cos B – sin A sin B)] = 2 + 2 cos (x + y) = 2 [1 + cos (x + y)] = 2[1 + \(2 \cos ^{2}\left(\frac{x+y}{2}\right)\) – 1] [∵ cos 2A = 2 cos 2  A – 1] = 4 c0s 2  \(\left(\frac{x+y}{2}\right)\) = R.H.S. Hence proved.

Miscellaneous Exercise Class 11 Maths Question 4: Prove that: (cos x – cos y) 2  + (sin x – sin y) 2  = 4 sin 2  \(\frac{x-y}{2}\) Ans:

L.H.S.= (cos x – cos y) 2  + (sin x – sin y) 2 = cos 2  x + cos 2  y – 2 cos x cos y + sin 2  x + sin 2  y – 2 sin x sin y = (cos 2  x + sin 2  x) + (cos 2  y + sin 2  y) – 2 [cos x cos y + sin x sin y] = 1 + 1 – 2 [cos (x – y)] = 2 [1 – {1 – 2 sin 2  \(\left(\frac{x-y}{2}\right)\)}] [∵ cos 2A = 1 – 2 sin 2  A] = 4 sin 2  \(\left(\frac{x-y}{2}\right)\) = R.H.S. Hence proved.

Miscellaneous Exercise Class 11 Maths Question 5: Prove that: sin x + sin 3x + sin 5x + sin 7x = 4 cos x cos 2x sin 4x Ans:

It is known that sin A + sin B = 2 \(\sin \left(\frac{A+B}{2}\right) \cdot \cos \left(\frac{A-B}{2}\right)\) ∴ L.H.S. = (sin x + sin 3x) + (sin 5x + sin 7x) = (sin x + sin 5x) + (sin 3x + sin 7x) = \(2 \sin \left(\frac{x+5 x}{2}\right)\) . \(\cos \left(\frac{x-5 x}{2}\right)+2 \sin \left(\frac{3 x+7 x}{2}\right) \cos \left(\frac{3 x-7 x}{2}\right)\) = 2 sin 3x cos (- 2x) + 2 sin 5x cos (- 2x) = 2 sin 3x cos 2x + 2 sin 5x cos 2x = 2 cos 2x [sin 3x + sin 5x] = 2 cos 2x [latex]2 \sin \left(\frac{3 x+5 x}{2}\right) \cdot \cos \left(\frac{3 x-5 x}{2}\right)[/latex] = 2 cos 2x [2 sin 4x . cos (- x)] = 4 cos 2x sin 4x cos x = R.H.S. Hence proved.

Miscellaneous Exercise Class 11 Maths Question 6: Prove that: \(\frac{(\sin 7 x+\sin 5 x)+(\sin 9 x+\sin 3 x)}{(\cos 7 x+\cos 5 x)+(\cos 9 x+\cos 3 x)}\) = tan 6x Ans: It is known that

प्रश्न 3. एक-पहिया एक मिनट में 360° परिक्रमण करता है तो एक सेकंड में कितने रेडियन माप का कोण बनाएगा? हल: परिक्रमण में पहिया द्वारा बना कोण = 27 रेडियन 360 परिक्रमण में पहिया द्वारा बना कोण = 360 x 2π रेडियन 1 मिनट अर्थात् 60 सेकण्ड में 360 x 2π रेडियन का कोण बनता है। 1 सेकण्ट में चहिया द्वारा बना कोण = \(\frac { 360\times 2\pi }{ 60 }\) = 12π रेडियन।

Exercise 3.1

Q.1: Calculate the radian measurement of the given degree measurement:

(ii). 240 ∘

(iii). − 47 ∘ 30 ‘

(iv). 520 ∘

Q.2: Calculate the degree measurement of the given degree measurement: [Use π = \(\\ \frac { 22 }{ 7 } \)]

(i) \(\\ \frac { 11 }{ 16 } \)

(iii) \(\frac { 5\pi }{ 3 } \)

(iv) \(\frac { 7\pi }{ 6 } \)

Q.3: In a minute, wheel makes 360 revolutions. Through how many radians does it turn in 1 second?

Q.4: Calculate the degree measurement of the angle subtended at the centre of a circle of radius 100 m by an arc of length 22 m.

Q.5: In a circle of diameter 40 m, the length of the chord 20 m. Find the length of minor arc of chord.

Q.6: In two circles, arcs which has same length subtended at an angle of 60 ∘ and 75 ∘ at the center. Calculate the ratio of their radii.

Q.7: Calculate the angle in radian through which a pendulum swings if the length is 75 cm and the tip describes an arc of length

(iii) 21 cm

Exercise 3.2

Q.1: Calculate the values of five trigonometric func. if cos y = \(– \frac { 1 }{ 2 } \) and y lies in 3 rd quadrant.

(iii) cosec y

Q.2: Calculate the other five trigonometric function if we are given the values for sin y = \(\\ \frac { 3 }{ 5 } \), where y lies in second quadrant.

Q.3: Find the values of other five trigonometric functions if c o t y =\(\\ \frac { 3 }{ 4 } \) , where y lies in the third quadrant.

Q.4: Find the values of other five trigonometric if s e c y =\(\\ \frac { 13 }{ 5 } \) , where y lies in the fourth quadrant.

Q.5: Find the values of other five trigonometric function if tan y = \(– \frac { 5 }{ 12 } \) and y lies in second quadrant.

Q.6: Calculate the value of trigonometric function sin 765°.

Q.7: Calculate the value of trigonometric function cosec [-1410°]

Q.8: Calculate the value of the trigonometric function tan \(\frac { 19\pi }{ 3 } \) .

Q.9: Calculate the value of the trigonometric function sin \(-\frac { 11\pi }{ 3 } \) .

Q.10: Calculate the value of the trigonometric function cot \(-\frac { 15\pi }{ 4 } \)

Exercise 3.3

Q.1: Prove:

sin²\(\frac { \pi }{ 6 } \) + cos² \(\frac { \pi }{ 3 } \) – tan² \(\frac { \pi }{ 4 } \) = \(– \frac { 1 }{ 2 } \)

Q.2: Prove:

2 sin² \(\frac { \pi }{ 6 } \) + c o s e c ² \(\frac { 7\pi }{ 6 } \) 6 cos ² \(\frac { \pi }{ 3 } \) =\(\\ \frac { 3 }{ 2 } \)

Q.3: Prove:

cot ² \(\frac { \pi }{ 6 } \) + c o s e c \(\frac { 5\pi }{ 6 } \) + 3 tan ² latex s=2]\frac { \pi }{ 6 } [/latex] = 6

Q.4: Prove:

2 sin ² \(\frac { 3\pi }{ 4 } \) + 2 cos ² \(\frac { \pi }{ 4 } \) + 2 sec ² \(\frac { \pi }{ 3 } \) = 10

Q.5: Calculate the value of:

(i). sin 75 ∘

(ii). tan 15 ∘

cos ( \(\frac { \pi }{ 4 } \) – x ) cos ( \(\frac { \pi }{ 4 } \) – y ) – sin ( \(\frac { \pi }{ 4 } \) – x ) sin ( \(\frac { \pi }{ 4 } \) – y ) = sin ( x + y )

Q.7: Prove:

\(\frac { tan(\frac { \pi }{ 4 } +x) }{ tan(\frac { \pi }{ 4 } -x) } ={ \left( \frac { 1+tanx }{ 1-tanx } \right) }^{ 2 }\)

Q.8: Prove:

\(\frac { cos(\pi +x)cos(-x) }{ sin(\pi -x)cos\left( \frac { \pi }{ 2 } +x \right) } ={ cot }^{ 2 }x\)

Q.9: Prove:

\(cos(\frac { 3\pi }{ 2 } +x)cos(2\pi +x)[cot(\frac { 3\pi }{ 2 } -x)+cot(2\pi +x)]=1\)

Q.10: Prove:

sin ( n + 1 ) x sin ( n + 2 ) x + cos ( n + 1 ) x cos ( n + 2 ) x = cos x

Q.11 Prove:

\(cos(\frac { 3\pi }{ 4 } +x)-cos(\frac { 3\pi }{ 4 } -x)\)= − √2 sin x

Q.12: Prove:

sin² 6 x – sin ² 4 x = sin2 x sin 10 x

Q.13: Prove:

cos ² 2 x – cos ² 6 x = sin 4 x sin 8 x

Q.14:Prove:

sin 2 x + 2 sin 4 x + sin 6 x = 4 cos ² x sin 4 x

Q.15: Prove:

cot 4 x ( sin 5 x + sin 3 x ) = cot x ( sin 5 x – sin 3 x )

Q.16: Prove:

\(\frac { cos9x-cos5x }{ sin17x-sin3x } =-\frac { sin2x }{ cos10x } \)

Q.17: Prove:

\(\frac { sin5x+sin3x }{ cos5x+cos3x } =tan4x\)

Q.18: Prove:

\(\frac { sinx-siny }{ cosx+cosy } =tan\frac { x-y }{ 2 } \)

Q.19: Prove:

\(\frac { sinx+sin3x }{ cosx+cos3x } =tan2x\)

Q.20: Prove:

\(\frac { sinx-sin3x }{ { sin }^{ 2 }x-{ cos }^{ 2 }x } =2sinx\)

Q.21: Prove:

\(\frac { cos4x+cos3x+cos2x }{ sin4x+sin3x+sin2x } =cot3x\)

Q.22: Prove:

cot x cot 2 x – cot 2 x cot 3 x – cot 3 x cot x = 1

Q.23: Prove:

\(tan4x=\frac { 4tanx(1-{ tan }^{ 2 }x) }{ 1-6{ tan }^{ 2 }x+{ tan }^{ 4 }x } \)

Q.24: Prove:

cos 4 x = 1 – 8 sin² x cos² x

Q.25: Prove:

cos 6 x = 32 cos 6 x – 48 cos 4 x + 18 cos 2 x − 1

Exercise 3.4

Q.1: Find general solutions and the principle solutions of the given equation: tan x = √3

Q.2: Find general solutions and the principle solutions of the given equation: sec x = 2

Q.3: Find general solutions and the principle solutions of the given equation: cot = − √3

Q.4: Find general solutions and the principle solutions of the given equation: cosec x = -2

Q.5: Find the general solution of the given equation: cos 4x = cos 2x

Q.6: Find the general solution of the given equation: cos 3x + cos x – cos 2x = 0

Q.7: Find the general solution of the given equation: sin 2x + cos x = 0

Q.8: Find the general solution of the given equation: sec² 2 x = 1 – tan 2 x

Q.9: Find the general solution of the given equation: sin x + sin 3x + sin 5x = 0

Miscellaneous Exercise

Q.1: Prove that:

\(2cos\frac { \pi }{ 13 } cos\frac { 9\pi }{ 13 } +cos\frac { 3\pi }{ 13 } +cos\frac { 5\pi }{ 13 } =0\)

Q.2: Prove that:

( sin 3 x + sin x ) sin x + ( cos 3 x – cos x ) cos x = 0

Q-3: Prove that:

( cos x + cos y )² + ( sin x – sin y ) ² = 4 cos ²\(\\ \frac { x+y }{ 2 } \)

Q-4: Prove that:

( cos x – cos y ) ² + ( sin x – sin y ) ² = 4 sin ² \(\\ \frac { x-y }{ 2 } \)

Q-5: Prove that:

sin x + sin 3 x + sin 5 x + sin 7 x = 4 cos x cos 2 x cos 4 x

Q-6: Prove that:

\(\frac { (sin7x+sin5x)+(sin9x+sin3x) }{ (cos7x+cos5x)+(cos9x+cos3x) } =tan6x\)

Q-7: Show that: sin 3 y + sin 2 y – sin y = 4 sin y cos\(\\ \frac { y }{ 2 } \) cos\(\\ \frac { 3y }{ 2 } \)

Q-8: The value of tan y =\(– \frac { 4 }{ 2 } \) where y in in 2 nd quadrant then find out the values of sin\(\\ \frac { y }{ 2 } \) , cos\(\\ \frac { y }{ 2 } \) a n d tan\(\\ \frac { y }{ 2 } \) .

Q-9: The value of cos y = \(– \frac { 1 }{ 3 } \) where y in in 3 rd quadrant then find out the values of sin \(\\ \frac { y }{ 2 } \) , cos \(\\ \frac { y }{ 2 } \) a n d tan \(\\ \frac { y }{ 2 } \) .

Q-10: The value of sin y = \(\\ \frac { 1 }{ 4 } \) where y in in 2 nd quadrant then find out the values of sin \(\\ \frac { y }{ 2 } \) , cos \(\\ \frac { y }{ 2 } \) a n d tan \(\\ \frac { y }{ 2 } \) .

NCERT Solutions for Class 11 Maths All Chapters

• Chapter 1 Sets
• Chapter 2 Relations and Functions
• Chapter 3 Trigonometric Functions
• Chapter 4 Principle of Mathematical Induction
• Chapter 5 Complex Numbers and Quadratic Equations
• Chapter 6 Linear Inequalities
• Chapter 7 Permutation and Combinations
• Chapter 8 Binomial Theorem
• Chapter 9 Sequences and Series
• Chapter 10 Straight Lines
• Chapter 11 Conic Sections
• Chapter 12 Introduction to Three Dimensional Geometry
• Chapter 13 Limits and Derivatives
• Chapter 14 Mathematical Reasoning
• Chapter 15 Statistics
• Chapter 16 Probability

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Trigonometry For Class 11

Trigonometry is one of the major topics in Maths subject. Trigonometry deals with the measurement of angles and sides of a triangle. Usually, trigonometry is considered for the right-angled triangle. Also, its functions are used to find out the length of the arc of a circle, which forms a section in the circle with a radius and its center point.

If we break the word trigonometry, ‘Tri’ is a Greek word which means ‘Three’, ‘Gon’ means ‘length’, and ‘metry’ means ‘measurement’. So basically, trigonometry is a study of triangles, which has angles and lengths on its side. Trigonometry basics consist of sine, cosine and tangent functions. Trigonometry for class 11 contains trigonometric functions, identities to solve complex problems more simply.

Trigonometry Formulas

Here, you will learn trigonometry formulas for class 11 and trigonometric functions of Sum and Difference of two angles and trigonometric equations.

Starting with the basics of Trigonometry formulas , for a right-angled triangle ABC perpendicular at B, having an angle θ, opposite to perpendicular (AB), we can define trigonometric ratios as;

Sin θ = P/H

Cos θ = B/H

Tan θ = P/B

Cot θ = B/P

Sec θ = H/B

Cosec θ = H/P

P = Perpendicular

H = Hypotenuse

Trigonometry Functions

Trigonometry functions are measured in terms of radian for a circle drawn in the XY plane. Radian is nothing but the measure of an angle, just like a degree. The difference between the degree and radian is;

Degree : If rotation from the initial side to the terminal side is (1/360)th of revolution, then the angle is said to measure 1 degree.

1 degree=60minutes

1 minute=60 second

Radian: If an angle is subtended at the center by an arc of length ‘l, the angle is measured as radian. Suppose θ is the angle formed at the center, then

θ = Length of the arc/radius of the circle.

Relation between Degree and Radian:

2π radian = 360 °

π radian = 180 °

Where π = 22/7

Learn more about the relation between degree and radian here.

Table for Degree and Radian relation

Earlier we have discussed of  trigonometric ratios for a degree, here we will write the table in terms of radians.

Trigonometry Table

Sign of Trigonometric Functions

sin(-θ) = -sin θ

cos(-θ) = cos θ

tan(-θ) = -tan θ

cot(-θ) = -cot θ

sec(-θ) = sec θ

cosec(-θ) = -cosec θ

Click here to know more about the sign of trigonometric functions .

Also, go through the table given below to understand the behaviour of trigonometric functions with respect to their values in different quadrants.

This behaviour can be observed from the trigonometry graphs .

Trigonometric Functions of Sum and Product of two angles

sin (x+y) = sin x cos y + cos x sin y

sin (x-y) = sin x cos y – cos x sin y

cos (x+y) = cos x cos y – sin x sin y

cos (x-y) = cos x cos y + sin x sin y

sin (π/2 – x) = cos x

cos (π/2 – x) = sin x

tan (x+y) = (tan x + tan y) /(1−tan x tan y)

tan (x-y) = (tan x − tan y)/(1 + tan x tan y)

cot (x+y) = (cot x cot y −1)/(cot y + cot x)

cot(x-y) = (cot x cot y + 1)/( cot y − cot x)

cos 2x = cos 2 x-sin 2 x = 2cos 2 x-1 = 1-2sin 2 x = (1-tan 2 x)/(1+tan 2 x)

sin 2x = 2sin x cos x= 2tan x/(1+ tan 2 x)

tan 2x = 2 tan x/(1-tan 2 x)

sin 3x = 3 sin x – 4 sin 3 x

cos 3x = 4 cos 3  x – 3 cos x

\(\begin{array}{l}cos\ x+ cos\ y=2\ cos{(\frac{x+y}{2})}\ cos{(\frac{x-y}{2})}\\ cos\ x – cos\ y = -2\ sin{(\frac{x+y}{2})}\ sin{(\frac{x-y}{2})}\\ sin\ x + sin\ y = 2\ sin{(\frac{x+y}{2})}\ cos{(\frac{x-y}{2})}\\ sin\ x – sin\ y = 2\ cos{(\frac{x+y}{2})}\ sin{(\frac{x-y}{2})}\end{array} \)

2 cos x cos y = cos (x+y) + cos (x-y)

2 sin x sin y = cos (x-y) – cos (x+y)

2 sin x cos y= sin (x+y) + sin (x-y)

2 cos x sin y = sin (x+y) – sin (x-y)

To solve the trigonometric questions for class 11, all these functions and formulas are used accordingly. By practising those questions, you can memorize the formulas as well.

Video Lessons

Basic trigonometric ratios & identities.

Sum, Difference & Allied Angles Formulae

Transformation of Graphs

Sum and Difference of Angles

Multiple & Sub-multiple Angles

Transformation Formulae & Conditional Identities

Solved Examples

Prove that sin(x+y)/ sin(x−y) = (tan x + tan y)/(tan x–tan y)

LHS = sin(x+y)/sin(x−y)

= (sin x cos y + cos x  sin y)/(sin x cos y − cos x sin y)

Dividing numerator and denominator by cos x cos y, we get

= (tan x + tan y)/(tan x–tan y) ———–Proved.

Find the value of cos (31π/3).

We know that the value of cos x repeats after the interval 2π.

Thus, cos (31π/3) = cos (10π + π/3)

= cos π/3 = 1/2

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Additional Trigonometry Related Articles For Class 11

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