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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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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

How should I study for my upcoming exams?

First, learn to sit for at least 2 hours at a stretch

Solve every question of NCERT by hand, without looking at the solution.

Solve NCERT Exemplar (if available)

Sit through chapter wise FULLY INVIGILATED TESTS

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

After Completing everything mentioned above, Sit for atleast 6 full syllabus TESTS.

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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

Important questions are marked, and Formula sheet is also provided. Click on an exercise or topic to begin.

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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

NCERT Solutions Class 11 Maths Chapter 3 Trigonometric Functions

NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions are formulated to explain the fundamental application of trigonometric formulas and the graphs of their functions. Trigonometry is an important part of Class 11 maths that involves studying various relations between sides and angles of triangles . Trigonometric Functions are applied to define these relations and have numerous applications in various other fields, including science and engineering. They are often used for basic geometric calculations and to explain numeric solutions. The knowledge of Trigonometric Functions is vital for subjects like astronomy and geography. NCERT Solutions Class 11 Maths Chapter 3 will offer the right foundational skills in students to study trigonometric relations and functions along with their practical applications.

Chapter 3 of Class 11 Maths will enable students to generalize the concept of trigonometric ratios to trigonometric functions. Learning about the properties of Trigonometric Functions and operations based on them is crucial for math studies. With the regular practice of Class 11 Maths NCERT Solutions Chapter 3, students will quickly master the solutions to equations using Trigonometric Functions. The wide range of problems and examples provided in these solutions are beneficial in promoting an in-depth understanding of concepts. To learn and practice with NCERT Solutions Chapter 3 Trigonometric Functions, download the exercises provided in the links below.

• NCERT Solutions Class 11 Maths Chapter 3 Ex 3.1
• NCERT Solutions Class 11 Maths Chapter 3 Ex 3.2
• NCERT Solutions Class 11 Maths Chapter 3 Ex 3.3
• NCERT Solutions Class 11 Maths Chapter 3 Ex 3.4
• NCERT Solutions Class 11 Maths Chapter 3 Miscellaneous Ex

NCERT Solutions for Class 11 Maths Chapter 3 PDF

Trigonometry sees the use of many formulas, theorems, and steps to solve the sums hence, ensuring that kids allot an ample amount of time for practice is very important. NCERT Solutions for Class 11 Maths Chapter 3 are proficiently modeled to support higher-level math learning. The comprehensive format of these solutions is highly reliable to enhance problem-solving skills in engaging ways. To learn and practice trigonometric functions with these solutions, click on the links of the pdf file given below.

☛ Download Class 11 Maths NCERT Solutions Chapter 3 Trigonometric Functions

NCERT Class 11 Maths Chapter 3   Download PDF

NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions

NCERT solutions Class 11 Maths Chapter 3 Trigonometric Functions are extremely beneficial in developing mathematical reasoning in students. With the help of these well-structured resources, students will gain a simplistic approach for efficient exam preparation. The carefully placed illustrations, questions, and examples present in these solutions are sufficient to enhance the fundamental math knowledge to excel in exams. Kids can easily plan their curriculum and revision format with the help of these solutions. To practice the exercise-wise NCERT Solutions Class 11 Maths Trigonometric Functions, try the links given below.

• Class 11 Maths Chapter 3 Ex 3.1 - 7 Questions
• Class 11 Maths Chapter 3 Ex 3.2 - 10 Questions
• Class 11 Maths Chapter 3 Ex 3.3 - 25 Questions
• Class 11 Maths Chapter 3 Ex 3.4 - 9 Questions
• Class 11 Maths Chapter 3 Miscellaneous Ex - 10 Questions

☛  Download Class 11 Maths Chapter 3 NCERT Book

Topics Covered: NCERT solutions Class 11 Maths Chapter 3 Trigonometric Functions cover some important topics, including an introduction to trigonometric ratios and identities, the measure of angles, signs, domain and range of trigonometric functions . The other important topic included in these solutions is the trigonometric solutions of the sum and difference of two angles using formulas .

Total Questions: Class 11 Maths Chapter 3 Trigonometric Functions has 51 questions in 4 exercises plus 10 questions in one miscellaneous exercise. These are primarily based on the representation of trigonometric functions, their applications, and formulas.

List of Formulas in NCERT Solutions Class 11 Maths Chapter 3

Memorizing important formulas and concepts is necessary to understand the terms and operations related to trigonometric functions clearly. NCERT Solutions Class 11 Maths Chapter 3 will provide detailed knowledge of all these along with their applications through interesting illustrations. Each concept in these solutions is well-explained with suitable examples and sample problems to promote an in-depth understanding of this topic. Formulas form an integral part of this lesson hence, creating a formulas sheet can be useful for students when they need to practice this topic. Some of the important terms, formulas, and concepts related to Trigonometric Functions explained in these solutions are listed below:

• Trigonometric Functions: Trigonometric Functions are real functions that relate the angle of a r ight-angled triangle to the ratio of the length of its sides. Sin, Cos, and Tan are the three primary functions.
• Trigonometric Equations: Trigonometric equations are equations that involve the use of trigonometric functions. These trigonometric equations are also known as trigonometric identities when the values of unknown angles for which the functions are defined are satisfied.
• Trigonometric Identities: A trigonometric equation that involves the sum and difference of angles represent a trigonometric identity. For example, sin 2 θ + cos 2 θ = 1
• Sum and Difference Identities: The sum and difference identities include the following formulas.
• sin(x+y) = sin(x)cos(y) + cos(x)sin(y)
• cos(x+y) = cos(x)cos(y) – sin(x)sin(y)
• tan(x+y) = (tan x + tan y)/ (1−tan x • tan y)
• sin(x–y) = sin(x)cos(y) – cos(x)sin(y)
• cos(x–y) = cos(x)cos(y) + sin(x)sin(y)
• tan(x−y) = (tan x–tan y)/ (1+tan x • tan y)

FAQs on NCERT Solutions Class 11 Maths Chapter 3

What is the importance of ncert solutions for class 11 maths chapter 3 trigonometric functions.

NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions are designed by well-qualified teachers and math experts to promote proficient math learning. Each exercise of these solutions is formulated as per the CBSE guidelines to support in-depth learning of Trigonometric Functions and their relations. These competently structured solutions are suitable to enhance math proficiency in students. Regular practice of these solutions will boost students’ confidence in scoring well.

What are the Important Topics Covered in NCERT Solutions Class 11 Maths Chapter 3?

NCERT Solutions Class 11 Maths Chapter 3 briefly introduces trigonometric ratios and identities along with some core concepts previously studied in grade 10. The important topics covered in these solutions are angle measures, trigonometric functions, their signs, domain, and range. The sum and difference of two angles using trigonometric functions are also included in this chapter.

Do I Need to Practice all Questions Provided in Class 11 Maths NCERT Solutions Trigonometric Functions?

NCERT Solutions Class 11 Maths Trigonometric Functions are designed in an effective way to facilitate the in-depth learning of concepts. Every question included in these solutions is aimed to improve the conceptual clarity for students to master them easily. The fundamental knowledge of Trigonometric Functions and their use will allow students to apply their knowledge in practical situations. These solutions also form the basis for studying advanced math topics, including calculus. Thus, all sums must be practiced with laser focus.

How Many Questions are there in Class 11 Maths NCERT Solutions Chapter 3 Trigonometric Functions?

NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions has 61 questions in 5 exercises. These problems are sufficient to provide a deep-seated understanding of the entire topic of trigonometric functions, including important formulas, identities, ratios, and operations related to Trigonometric Functions. The jargon-free and lucid math vocabulary used in these solutions are quite proficient in easily imparting a clear step-by-step understanding of each topic as well as subtopics.

What are the Important Formulas in Class 11 Maths NCERT Solutions Chapter 3?

NCERT Solutions Class 11 Maths Chapter 3 explains the formulas related to the sum and difference of two angles in trigonometric functions. Some of the important concepts related to this topic are based on deriving expressions for trigonometric identities of the sum and difference of angles. These vital concepts are described comprehensively with the help of interesting examples for students to grasp them better. Additionally, kids need to also revise the formulas they have encountered in previous classes as those are also necessary to attempt class 11 trigonometry problems.

Why Should I Practice NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions?

NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions is a reliable source of learning that gives full guidance for excellent exam preparation. With the help of these solutions, students can cover the whole syllabus of CBSE Class 11 Maths Chapter 3. The format of these solutions is quite intuitive to promote simple and easy learning. Thus, to get the best results kids need to practice these solutions.

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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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Unit 3: Trigonometric functions

Degrees and radians.

• Intro to radians (Opens a modal)
• Radians & degrees (Opens a modal)
• Degrees to radians (Opens a modal)
• Radians to degrees (Opens a modal)
• Radian angles & quadrants (Opens a modal)
• Radians & degrees Get 3 of 4 questions to level up!

Unit circle

• Unit circle (Opens a modal)
• The trig functions & right triangle trig ratios (Opens a modal)
• Trig values of π/4 (Opens a modal)
• Trig unit circle review (Opens a modal)
• Plotting angles on unit circle Get 3 of 4 questions to level up!
• Sign of trigonometric functions Get 3 of 4 questions to level up!
• Trig values of special angles Get 3 of 4 questions to level up!

Trigonometric functions

• Graph of y=sin(x) (Opens a modal)
• Graph of y=tan(x) (Opens a modal)
• Intersection points of y=sin(x) and y=cos(x) (Opens a modal)
• Find value of other trigonometric functions from given trigonometric function Get 3 of 4 questions to level up!

Trigonometric identities: Symmetry

• Sine & cosine identities: symmetry (Opens a modal)
• Tangent identities: symmetry (Opens a modal)
• Sine & cosine identities: periodicity (Opens a modal)
• Tangent identities: periodicity (Opens a modal)

Trigonometric identities: Sum and difference

• Trig angle addition identities (Opens a modal)
• Using the cosine angle addition identity (Opens a modal)
• Using the cosine double-angle identity (Opens a modal)
• Proof of the sine angle addition identity (Opens a modal)
• Proof of the cosine angle addition identity (Opens a modal)
• Trig challenge problem: cosine of angle-sum (Opens a modal)
• Trigonometric functions of sum and difference of angles Get 3 of 4 questions to level up!
• Sum and difference of trigonometric functions Get 3 of 4 questions to level up!
• Evaluate trigonometric expressions (intermediate) Get 3 of 4 questions to level up!

Trigonometric equations

• Proof of the Pythagorean trig identity (Opens a modal)
• Using the Pythagorean trig identity (Opens a modal)
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Trigonometric Functions Class 11 Notes CBSE Maths Chapter 3 [Free PDF Download]

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Revision Notes for CBSE Class 11 Maths Chapter 3 (Trigonometric Functions) - Free PDF Download

Trigonometry Functions are one of the most important topics in Class 11 Mathematics. It tells about the relationship between the sides and angles of a right-angle triangle. Trigonometric Functions Class 11 Notes are the important study material for the students looking to clear their basic concepts of trigonometric functions. It helps them to more confidently solve the trigonometry based questions as the Class 11 Maths Chapter 3 notes enable students to get an immediate overview of all the topics along with the formulas covered in Class 11 Maths Chapter 3.

These Trigonometric Functions Class 11 Notes are arranged systematically for students' comfort and students are suggested to refer to these revision notes whenever required. Students can prepare the complete chapter effectively with the help of Class 11 Maths notes Chapter 3 notes as the contents in these notes are made quite interactive.

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Trigonometric Functions Class 11 Notes Maths - Basic Subjective Questions

Section–A (1 Mark Questions)

1. If $tan\Theta =\frac{-4}{3}$ , then find $sin\Theta$.

Ans . Since, $\tan \theta=-\frac{4}{3}$ is negative, $\theta$ lies either in second quadrant or in fourth quadrant.

$$ \begin{aligned} & \because 1+\cot ^2 \theta=\operatorname{cosec}^2 \theta \\ & \Rightarrow 1+\frac{1}{\left(-\frac{4}{3}\right)^2}=\frac{1}{\sin ^2 \theta} \\ & \Rightarrow \sin ^2 \theta=\left(\frac{4}{5}\right)^2 \end{aligned} $$

Thus $\sin \theta=\frac{4}{5}$ if $\theta$ lies in the second quadrant or $\sin \theta=-\frac{4}{5}$, if $\theta$ lies in the fourth quadrant.

2. Find the greatest value of sin x cos x .

Ans. $\sin x \cos x=\frac{1}{2}(2 \sin x \cos x)=\frac{1}{2} \sin 2 x$ Greatest value of $\sin 2 x=1$

$\therefore$ Greatest value of $\dfrac{1}{2} \sin 2 x=\dfrac{1}{2} \times 1=\dfrac{1}{2}$

3. Find the degree measure of angle $\left ( \frac{\pi }{8} \right )^{c}$  radian.

Ans. We know that, $\pi$ radians $=180^{\circ}$

$$ \begin{aligned} & \Rightarrow \frac{\pi}{8} \text { radians }=22.5^{\circ} \\ & =22^{\circ}+0.5^{\circ}=22^{\circ}+30^{\prime}=22^{\circ} 30^{\prime} \end{aligned} $$

4. Evaluate: $sin \left ( \frac{-15\pi }{4} \right )$ .

$$ \begin{aligned} & \text { 4. } \sin \left(\frac{-15 \pi}{4}\right)=\sin \left(\frac{-16 \pi+\pi}{4}\right) \\ & =\sin \left(-4 \pi+\frac{\pi}{4}\right) \\ & =\sin \left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}} \end{aligned} $$

5. The minute hand of a watch is 1.5 cm long. How far does it tip move in 40 minutes? (Use $\pi$ = 3.14) .

Ans. Angle rotated by minute hand in 60 minutes = $2\pi$     radians 

Therefore, angle rotated by minute hand in 40 minutes = $\frac{40}{60}\times2\pi =\frac{4\pi }{3}$  radians.

Hence, the required distance travelled is given by

$l=r\Theta =1.5\times\frac{4\pi }{3}cm=2\pi\;cm$

$=2\times3.14\;cm=6.28\;cm$

Section–B (2 Marks Questions)

6. Prove that cot2x cotx - cot3x cot2x - cot3x cotx = 1

Ans. We have, cot 3x=cot (2x+x)

$\Rightarrow cot\;3x=\dfrac{cot\;2x\;cot\;x -1}{cot\;2x+cot\;x}$

$\Rightarrow$ Cot 3x cot 2x + cot 3x cot x = cot 2x cot x-1

$\Rightarrow$ cot 2x cot x - cot 3x cot 2x - cot 3x xot x=1

7. Evaluate: $\frac{1-\tan^2{15^\circ}}{1+\tan^2{15^\circ}}$ 

Ans. Given that: $\frac{1-\tan^2{15^\circ}}{1+\tan^2{15^\circ}}$

Let $\Theta =15^{\circ}$

$\therefore 2\Theta =30^{\circ}$

We know that, $\therefore cos\;2\Theta =\frac{1-\tan^2{15^\circ}}{1+\tan^2{15^\circ}}$

$\Rightarrow  cos\;30 =\frac{1-\tan^2{15^\circ}}{1+\tan^2{15^\circ}}$

$\frac{1-\tan^2{15^\circ}}{1+\tan^2{15^\circ}}=\frac{\sqrt{3}}{3}$

8. If for real value of x , $cos\Theta =x+\frac{1}{x}$ then what can you say about $\Theta$    ?

Ans. Given that: $cos\Theta =x+\frac{1}{x}$ 

$\Rightarrow cos\Theta =x+\frac{1}{x}$

$$ \begin{aligned} & \Rightarrow x^2+1=x \cos \theta \\ & \Rightarrow x^2-x \cos \theta+1=0 \end{aligned} $$

For real value of $x, b^2-4 a c \geq 0$

$$ \begin{aligned} & \Rightarrow(-\cos \theta)^2-4 \times 1 \times 1 \geq 0 \\ & \Rightarrow \cos ^2 \theta-4 \geq 0 \\ & \Rightarrow \cos ^2 \theta \geq 4 \\ & \Rightarrow \cos \theta \geq \pm 2 \quad[\because-1 \leq \cos \theta \leq 1] \end{aligned} $$

So, the value of $\theta$ is not possible.

9. Find the value of $sin\left ( \frac{\pi }{4}+\Theta  \right )-cos \left ( \frac{\pi }{4}-\Theta  \right )$ .

Ans. Given expression :

$$ \begin{aligned} & \sin \left(\frac{\pi}{4}+\theta\right)-\cos \left(\frac{\pi}{4}-\theta\right) \\ & \sin \left(\frac{\pi}{4}+\theta\right)=\sin \left(\frac{\pi}{4}\right) \cos \theta+\cos \left(\frac{\pi}{4}\right) \sin \theta \\ & =\frac{1}{\sqrt{2}} \cos \theta+\frac{1}{\sqrt{2}} \sin \theta \\ & \cos \left(\frac{\pi}{4}-\theta\right)=\cos \left(\frac{\pi}{4}\right) \cos \theta+\sin \left(\frac{\pi}{4}\right) \sin \theta \\ & =\frac{1}{\sqrt{2}} \cos \theta+\frac{1}{\sqrt{2}} \sin \theta \\ & \therefore \sin \left(\frac{\pi}{4}+\theta\right)-\cos \left(\frac{\pi}{4}-\theta\right) \\ & =\frac{1}{\sqrt{2}} \cos \theta+\frac{1}{\sqrt{2}} \sin \theta-\frac{1}{\sqrt{2}} \cos \theta-\frac{1}{\sqrt{2}} \sin \theta \\ & =0 \end{aligned} $$

10. Find the value of $tan\;\frac{\pi }{12}$ .

Ans. $\tan \frac{\pi}{12}=\tan \left(\frac{\pi}{4}-\frac{\pi}{6}\right)$

$$ \begin{aligned} & =\frac{\tan \frac{\pi}{4}-\tan \frac{\pi}{6}}{1+\tan \frac{\pi}{4} \cdot \tan \frac{\pi}{6}}=\frac{1-\frac{1}{\sqrt{3}}}{1+1 \cdot \frac{1}{\sqrt{3}}}=\frac{\sqrt{3}-1}{\sqrt{3}+1} \\ & =\frac{(\sqrt{3}-1)(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}=\frac{4-2 \sqrt{3}}{2}=2-\sqrt{3} \end{aligned} $$

11. Prove that : $tan\;225^{\circ}\;cot\;405^{\circ}+tan\;765^{\circ}\;cot\;675^{\circ}=0$ 

$$ \begin{aligned} & \text { 11. LHS }= \\ & \tan 225^{\circ} \cot 405^{\circ}+\tan 765^{\circ} \cot 675^{\circ} \\ & =\tan \left(180^{\circ}+45^{\circ}\right) \cot \left(360^{\circ}+45^{\circ}\right) \\ & \quad \quad \quad \quad \tan \left(360^{\circ} \times 2+45^{\circ}\right) \cot \left(360^{\circ} \times 2-45^{\circ}\right) \\ & =\tan 45^{\circ} \cot 45^{\circ}+\tan 45^{\circ}\left[-\cot 45^{\circ}\right] \\ & =1 \times 1-1 \times 1 \\ & =0=\text { RHS } \end{aligned} $$

Hence proved.

12. For $0<x<\frac{\pi }{2}$ , show that $\sqrt{\frac{(1-cos2x)}{1+cos2x}}=tan\;x$ 

$$ \begin{aligned} & \text { LHS }=\sqrt{\frac{(1-\cos 2 x)}{(1+\cos 2 x)}} \\ & =\sqrt{\frac{1-\left(1-2 \sin ^2 x\right)}{1+\left(2 \cos ^2 x-1\right)}}=\sqrt{\frac{2 \sin ^2 x}{2 \cos ^2 x}} \\ & =|\tan x|=\tan x \quad\left\{\because 0<x<\frac{\pi}{2}\right\} \\ & =\text { RHS } \end{aligned} $$

13. Prove that: $\frac{cos(2\pi +x)cosec(2\pi +x)tan\left ( \frac{\pi }{2}+x \right )}{sec\left ( \frac{\pi }{2}+x \right )cos\;x\;cot(\pi +x)}=1$ 

Ans. LHS $=$

$$\cos (2 \pi+x) \operatorname{cosec}(2 \pi+x) \tan \left(\frac{\pi}{2}+x\right)$$

$$ \begin{aligned} & \sec \left(\frac{\pi}{2}+x\right) \cos x \cot (\pi+x) \\ \Rightarrow & \frac{-\cos x \operatorname{cosec} x \cot x}{-\operatorname{cosec} x \cos x \cot x}=1 \end{aligned} $$

Hence proved

PDF Summary - Class 11 Maths Trigonometric Functions Notes (Chapter 3)

1. The Meaning of Trigonometry

${{\text{Tri }}}{{\text{ Gon }}}{{\text{ Metron }}}$

$\downarrow\quad\;\;\;\;\downarrow\quad\;\;\;\;\downarrow$

$3\quad{{\text{ sides }}}{{\text{ Measure }}}$

As a result, this area of mathematics was established in the ancient past to measure a triangle's three sides, three angles, and six components. Time-trigonometric functions are utilised in a variety of ways nowadays. The sine and cosine of an angle in a right-angled triangle are the two fundamental functions, and there are four more derivative functions.

2. Basic Trigonometric Identities

(a) ${\sin ^2}\theta  + {\cos ^2}\theta  = 1: - 1 \leqslant \sin \theta  \leqslant 1; - 1 \leqslant \cos \theta  \leqslant 1\forall \theta  \in {\text{R}}$

(b) ${\sec ^2}\theta  - {\tan ^2}\theta  = 1:|\sec \theta | \geqslant 1\forall \theta  \in {\text{R}}$

(c) ${\operatorname{cosec} ^2}\theta  - {\cot ^2}\theta  = 1:|\operatorname{cosec} \theta | \geqslant 1\forall \theta  \in {\text{R}}$

Trigonometric Ratios of Standard Angles:

The relation between these trigonometric identities with the sides of the triangles can be given as follows:

Sine θ $=$ Opposite/Hypotenuse

Cos θ  $=$ Adjacent/Hypotenuse

Tan θ  $=$ Opposite/Adjacent

Cot θ $=$ Adjacent/Opposite

Cosec θ  = Hypotenuse/Opposite

Sec θ  = Hypotenuse/Adjacent

The following are the signs of trigonometric ratios in different quadrants:

3. Trigonometric Ratios of Allied Angles

We might calculate the trigonometric ratios of angles of any value using the trigonometric ratio of allied angles.

1. Sin(–θ)=–Sinθ

2. Cos(–θ)=Cosθ

3. Tan(–θ)=–Tanθ

4. Sin(90 o –θ)=Cosθ

5. Cos(90 o –θ)=Sinθ

6. Tan(90 o –θ)=Cotθ

7. Sin(180 o –θ)=Sinθ

8. Cos(180 o –θ)=–Cosθ

9. Tan(180 o –θ)=–Tanθ

10. Sin(270 o –θ)=–Cosθ

11. Cos(270 o –θ)=–Sinθ

12. Tan(270 o –θ)=Cotθ

13. Sin(90 o +θ)=Cosθ

14. Cos(90 o +θ)=–Sinθ

15. Tan(90 o +θ)=–Cotθ

16. Sin(180 o +θ)=–Sinθ

17. Cos(180 o +θ)=–Cosθ

18. Tan(180 o +θ)=Tanθ

19. Sin(270 o +θ)=–Cosθ

20. Cos(270 o +θ)=Sinθ

21. Tan(270 o +θ)=–Cotθ

4. Trigonometric Functions of Sum or Difference of Two Angles

(a) $\sin ({\text{A}} + {\text{B}}) = \sin {\text{A}}\cos {\text{B}} + \cos {\text{A}}\sin {\text{B}}$

(b) $\sin ({\text{A}} - {\text{B}}) = \sin {\text{A}}\cos {\text{B}} - \cos {\text{A}}\sin {\text{B}}$

(c) $\cos (A + B) = \cos A\cos B - \sin A\sin B$

(d) $\cos ({\text{A}} - {\text{B}}) = \cos {\text{A}}\cos {\text{B}} + \sin {\text{A}}\sin {\text{B}}$

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

(f) $\tan ({\text{A}} - {\text{B}}) = \dfrac{{\tan {\text{A}} - \tan {\text{B}}}}{{1 + \tan {\text{A}}\tan {\text{B}}}}$

(g) $\cot (A + B) = \dfrac{{\cot A\cot B - 1}}{{\cot B + \cot A}}$

(f) $\cot ({\text{A}} - {\text{B}}) = \dfrac{{\cot {\text{A}}\cot {\text{B}} + 1}}{{\cot {\text{B}} - \cot {\text{A}}}}$

(h) ${\sin ^2}\;{\text{A}} - {\sin ^2}\;{\text{B}} = {\cos ^2}\;{\text{B}} - {\cos ^2}\;{\text{A}} = \sin ({\text{A}} + {\text{B}}) \cdot \sin ({\text{A}} - {\text{B}})$

(i) ${\cos ^2}\;{\text{A}} - {\sin ^2}\;{\text{B}} = {\cos ^2}\;{\text{B}} - {\sin ^2}\;{\text{A}} = \cos ({\text{A}} + {\text{B}}) \cdot \cos ({\text{A}} - {\text{B}})$

(j) $\tan (A + B + C) = \dfrac{{\tan A + \tan B + \tan C - \tan A\tan B\tan C}}{{1 - \tan A\tan B - \tan B\tan C - \tan C\tan A}}$

5. Multiple Angles and Half Angles

(a) $\sin 2\;{\text{A}} = 2\sin {\text{A}}\cos {\text{A}};\quad \sin \theta  = 2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}$

(b) $\cos 2\;{\text{A}} = {\cos ^2}\;{\text{A}} - {\sin ^2}\;{\text{A}} = 2{\cos ^2}\;{\text{A}} - 1 = 1 - 2{\sin ^2}\;{\text{A}}$  

$2{\cos ^2}\dfrac{\theta }{2} = 1 + \cos \theta ,2{\sin ^2}\dfrac{\theta }{2} = 1 - \cos \theta $

(c) $\tan 2\;{\text{A}} = \dfrac{{2\tan {\text{A}}}}{{1 - {{\tan }^2}\;{\text{A}}}};\tan \theta  = \dfrac{{2\tan \dfrac{\theta }{2}}}{{1 - {{\tan }^2}\dfrac{\theta }{2}}}$

(d) $\sin 2\;{\text{A}} = \dfrac{{2\tan {\text{A}}}}{{1 - {{\tan }^2}\;{\text{A}}}};\cos 2\;{\text{A}} = \dfrac{{1 - {{\tan }^2}\;{\text{A}}}}{{1 + {{\tan }^2}\;{\text{A}}}}$

(e) $\sin 3\;{\text{A}} = 3\sin {\text{A}} - 4{\sin ^3}\;{\text{A}}$

(f) $\cos 3\;{\text{A}} = 4{\cos ^3}\;{\text{A}} - 3\cos {\text{A}}$

(g) $\tan 3\;{\text{A}} = \dfrac{{3\tan {\text{A}} - {{\tan }^3}\;{\text{A}}}}{{1 - 3{{\tan }^2}\;{\text{A}}}}$

6. Transformation of Products into Sum or Difference of Sines & Cosines

(a) $2\sin {\text{A}}\cos {\text{B}} = \sin ({\text{A}} + {\text{B}}) + \sin ({\text{A}} - {\text{B}})$

(b) $2\cos {\text{A}}\sin {\text{B}} = \sin ({\text{A}} + {\text{B}}) - \sin ({\text{A}} - {\text{B}})$

(c) $2\cos {\text{A}}\cos {\text{B}} = \cos ({\text{A}} + {\text{B}}) + \cos ({\text{A}} - {\text{B}})$

(d) $2\sin {\text{A}}\sin {\text{B}} = \cos ({\text{A}} - {\text{B}}) - \cos ({\text{A}} + {\text{B}})$

7. Factorisation of the Sum or Difference of Two Sines or Cosines

(a) $\sin {\text{C}} + \sin {\text{D}} = 2\sin \dfrac{{{\text{C}} + {\text{D}}}}{2}\cos \dfrac{{{\text{C}} - {\text{D}}}}{2}$

(b) $\sin {\text{C}} - \sin {\text{D}} = 2\cos \dfrac{{{\text{C}} + {\text{D}}}}{2}\sin \dfrac{{{\text{C}} - {\text{D}}}}{2}$

(c) $\cos C + \cos D = 2\cos \dfrac{{C + D}}{2}\cos \dfrac{{C - D}}{2}$

(d) $\cos {\text{C}} - \cos {\text{D}} =  - 2\sin \dfrac{{{\text{C}} + {\text{D}}}}{2}\sin \dfrac{{{\text{C}} - {\text{D}}}}{2}$

8. Important Trigonometric Ratios

(a) $\sin {\text{n}}\pi  = 0;\cos {\text{n}}\pi  = {( - 1)^{\text{n}}};\tan {\text{n}}\pi  = 0$ where ${\text{n}} \in {\text{Z}}$

(b) $\sin {15^\circ }$ or $\sin \dfrac{\pi }{{12}} = \dfrac{{\sqrt 3  - 1}}{{2\sqrt 2 }} = \cos {75^\circ }$ or $\cos \dfrac{{5\pi }}{{12}}$ ; 

$\cos {15^\circ }$ or $\cos \dfrac{\pi }{{12}} = \dfrac{{\sqrt 3  + 1}}{{2\sqrt 2 }} = \sin {75^\circ }$ or $\sin \dfrac{{5\pi }}{{12}}$  

$\tan {15^\circ } = \dfrac{{\sqrt 3  - 1}}{{\sqrt 3  + 1}} = 2 - \sqrt 3  = \cot {75^\circ }$  

$\tan {75^\circ } = \dfrac{{\sqrt 3  + 1}}{{\sqrt 3  - 1}} = 2 + \sqrt 3  = \cot {15^\circ }$

(c) $\sin \dfrac{\pi }{{10}}$ or $\sin {18^\circ } = \dfrac{{\sqrt 5  - 1}}{4}$ & $\cos {36^\circ }$ or $\cos \dfrac{\pi }{5} = \dfrac{{\sqrt 5  + 1}}{4}$

9. Conditional Identities

If ${\text{A}} + {\text{B}} + {\text{C}} = \pi $ then :

(i) $\sin 2\;{\text{A}} + \sin 2\;{\text{B}} + \sin 2{\text{C}} = 4\sin {\text{A}}\sin {\text{B}}\sin {\text{C}}$

(ii) $\sin {\text{A}} + \sin {\text{B}} + \sin {\text{C}} = 4\cos \dfrac{{\text{A}}}{2}\cos \dfrac{{\text{B}}}{2}\cos \dfrac{{\text{C}}}{2}$

(iii) $\cos 2\;{\text{A}} + \cos 2\;{\text{B}} + \cos 2{\text{C}} =  - 1 - 4\cos {\text{A}}\cos {\text{B}}\cos {\text{C}}$

(iv) $\cos {\text{A}} + \cos {\text{B}} + \cos {\text{C}} = 1 + 4\sin \dfrac{{\text{A}}}{2}\sin \dfrac{{\text{B}}}{2}\sin \dfrac{{\text{C}}}{2}$

(v) $\tan {\text{A}} + \tan {\text{B}} + \tan {\text{C}} = \tan {\text{A}}\tan {\text{B}}\tan {\text{C}}$

(vi) $\tan \dfrac{{\text{A}}}{2}\tan \dfrac{{\text{B}}}{2} + \tan \dfrac{{\text{B}}}{2}\tan \dfrac{{\text{C}}}{2} + \tan \dfrac{{\text{C}}}{2}\tan \dfrac{{\text{A}}}{2} = 1$

(vii) $\cot \dfrac{{\text{A}}}{2} + \cot \dfrac{{\text{B}}}{2} + \cot \dfrac{{\text{C}}}{2} = \cot \dfrac{{\text{A}}}{2} \cdot \cot \dfrac{{\text{B}}}{2} \cdot \cot \dfrac{{\text{C}}}{2}$

(viii) $\cot A\cot B + \cot B\cot C + \cot C\cot A = 1$

10. Range of Trigonometric Expression

${{\text{E}} = {\text{a}}\sin \theta  + {\text{b}}\cos \theta }$

${{\text{E}} = \sqrt {{{\text{a}}^2} + {{\text{b}}^2}} \sin (\theta  + \alpha ),\left( {{\text{ where }}\tan \alpha  = \dfrac{{\text{b}}}{{\text{a}}}} \right)}$ 

${{\text{E}} = \sqrt {{{\text{a}}^2} + {{\text{b}}^2}} \cos (\theta  - \beta ),\left( {{\text{ where }}\tan \beta  = \dfrac{{\text{a}}}{{\text{b}}}} \right)}$

Hence for any real value of $\theta , - \sqrt {{a^2} + {b^2}}  \leqslant E \leqslant \sqrt {{a^2} + {b^2}} $

The trigonometric functions are very important for studying triangles, light, sound or waves. The values of these trigonometric functions in different domains and ranges can be used from the following table:

Trigonometric Functions in Different Domains and Ranges

11. Sine and Cosine Series

(a) $\quad \sin \alpha  + \sin (\alpha  + \beta ) + \sin (\alpha  + 2\beta ) +  \ldots . + \sin (\alpha  + \overline {n - 1} \beta )$

$ = \dfrac{{\sin \dfrac{{{\text{n}}\beta }}{2}}}{{\sin \dfrac{\beta }{2}}}\sin \left( {\alpha  + \dfrac{{{\text{n}} - 1}}{2}\beta } \right)$

${\cos \alpha  + \cos (\alpha  + \beta ) + \cos (\alpha  + 2\beta ) +  \ldots  + \cos (\alpha  + \overline {n - 1} \beta )}$

${ = \dfrac{{\sin \dfrac{{n\beta }}{2}}}{{\sin \dfrac{\beta }{2}}}\cos \left( {\alpha  + \dfrac{{n - 1}}{2}\beta } \right)}$ 

12. Graphs of Trigonometric Functions 

(a). ${y = \sin x,}$

${x \in R;y \in [ - 1,1]}$ 

(b). $y = \cos x$

$x \in R;y \in [ - 1,1]$ 

(c) ${y = \tan x}$

${x \in R - \left\{ {(2n + 1)\dfrac{\pi }{2};n \in Z} \right\};y \in R}$ 

(d) ${y = \cot x}$

${x \in R - \{ n\pi ;n \in z\} ;y \in R}$ 

(e) ${y = \operatorname{cosec} x}$ 

${x \in R - \{ n\pi ;n \in Z\} ;y \in ( - \infty , - 1] \cup [1,\infty )}$

(f) $y = \sec x\quad $

$x \in R - \left\{ {(2n + 1)\dfrac{\pi }{2};n \in Z} \right\};y \in ( - \infty , - 1] \cup [1,\infty )$ 

Trigonometric Equations

13. Trigonometric Equations

Trigonometric equations are equations using trigonometric functions with unknown angles. 

e.g.,  $\cos \theta  = 0,{\cos ^2}\theta  - 4\cos \theta  = 1$ .

The value of the unknown angle that satisfies a trigonometric equation is called a solution.

e.g., $\quad \sin \theta  = \dfrac{1}{{\sqrt 2 }} \Rightarrow \theta  = \dfrac{\pi }{4}$ or $\theta  = \dfrac{\pi }{4},\dfrac{{3\pi }}{4},\dfrac{{9\pi }}{4},\dfrac{{11\pi }}{4}, \ldots $

As a result, the trigonometric equation can have an unlimited number of solutions and is categorised as follows:

(i). Principal Solution

As we know, the values of $\sin x$ and $\cos x$ will get repeated after an interval of $2 \pi$. In the same way, the values of $\tan x$ will get repeated after an interval of $\pi$. 

So, if the equation has a variable $0 \leq \mathrm{x}<2 \pi$, then the solutions will be termed as principal solutions. 

Find the principal solutions of the equation $\sin x=\dfrac{\sqrt{3}}{2}$.

Solution: We know that, $\sin \dfrac{\pi}{3}=\dfrac{\sqrt{3}}{2}$

Also, $\sin \dfrac{2 \pi}{3}=\sin \left(\pi-\dfrac{\pi}{3}\right)$

Now, we know that $\sin (\pi-x)=\sin x$. 

Hence, $\sin \dfrac{2 \pi}{3}=\sin \dfrac{\pi}{3}=\dfrac{\sqrt{3}}{2}$

Therefore, the principal solutions of $\sin x=\dfrac{\sqrt{3}}{2}$ are $\mathrm{x}=\dfrac{\pi}{3}$ and $\dfrac{2 \pi}{3}$.

(ii). General solution

A general solution is one that involves the integer 'n' and yields all trigonometric equation solutions. Also, the character ' $\mathrm{Z}$ ' is used to denote the set of integers.

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

Solution: We know that $\sin \dfrac{\pi}{3}=\dfrac{\sqrt{3}}{2}$. Therefore, $\sin x=-\dfrac{\sqrt{3}}{2}=-\sin \dfrac{\pi}{3}$

Using the unit circle properties, we get $\sin x=-\sin \dfrac{\pi}{3}=\sin \left(\pi+\dfrac{\pi}{3}\right)=\sin \dfrac{4 \pi}{3}$ Hence, $\sin x=\sin \dfrac{4 \pi}{3}$

Since, we know that for any real numbers $x$ and $y, \sin x=\sin y$ implies $x=n \pi+(-1)^{n} y$, where $n \in Z$.

So, we get, $x=n \pi+(-1)^{\mathrm{n}}\left(\dfrac{4 \pi}{3}\right)$

14.1 Results

1. $\quad \sin \theta  = 0 \Leftrightarrow \theta  = \operatorname{n} \pi $

2. $\cos \theta  = 0 \Leftrightarrow \theta (2{\text{n}} + 1)\dfrac{\pi }{2}$

3. $\tan \theta  = 0 \Leftrightarrow \theta  = {\text{n}}\pi $

4. $\sin \theta  = \sin \alpha  \Leftrightarrow \theta  = {\text{n}}\pi  + {( - 1)^{\text{n}}}\alpha $ , where $\alpha  \in \left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$

5. $\cos \theta  = \cos \alpha  \Leftrightarrow \theta  = 2{\text{n}}\pi  \pm \alpha $ , where $\alpha  \in [0,\pi ]$

6. $\tan \theta  = \tan \alpha  \Leftrightarrow \theta  = {\text{n}}\pi  + \alpha $ , where $\alpha  \in \left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$

7. ${\sin ^2}\theta  = {\sin ^2}\alpha  \Leftrightarrow \theta  = {\text{n}}\pi  \pm \alpha $ .

8. ${\cos ^2}\theta  = {\cos ^2}\alpha  \Leftrightarrow \theta  = $ n $\pi  \pm \alpha $ .

9. ${\tan ^2}\theta  = {\tan ^2}\alpha  \Leftrightarrow \theta  = {\text{n}}\pi  \pm \alpha $ .

10. $\sin \theta  = 1 \Leftrightarrow \theta  = (4{\text{n}} + 1)\dfrac{\pi }{2}$

11. $\cos \theta  = 1 \Leftrightarrow \theta  = 2{\text{n}}\pi $

12. $\cos \theta  =  - 1 \Leftrightarrow \theta  = (2{\text{n}} + 1)\pi $ .

13. $\sin \theta  = \sin \alpha $ and $\cos \theta  = \cos \alpha  \Leftrightarrow \theta  = 2{\text{n}}\pi  + \alpha $

Steps to Solve Trigonometric Functions:

The following are the stages of solving trigonometric equations:

Step 1: Decompose the trigonometric equation into a single trigonometric ratio, preferably the sine or cos function.

Step 2: Factor the trigonometric polynomial given in terms of the ratio.

Step 3: Write down the general solution after solving for each factor.

1. Unless otherwise stated, is treated as an integer throughout this chapter.

2. Unless the answer is required in a specific interval or range, the general solution should be supplied.

3. The angle's main value is regarded as $\alpha $ . (The main value is the angle with the least numerical value.)

Download Free Trigonometric Functions Class 11 Notes PDF

Class 11 students can easily download Trigonometric Functions Class 11 notes pdf and revise them at any time and anywhere. Chapter 3 Maths Class 11 pdf enables students to have excellent study patterns with which they can score maximum marks in the trigonometry section and even enjoy studying the topic.

Class 11 Maths Notes Chapter 3 summarize key points of the entire chapter in the easiest way so that students can remember them and also enhance their confidence before attempting the trigonometric questions in the exam. These revision notes are very important and beneficial as it helps you to quickly and effectively revise the entire chapter before the exam. Class 11 Maths Trigonometric Functions notes are prepared concisely from the latest edition of Class 11 Maths NCERT textbook.

The subject excerpts at Vedantu who prepared these revision notes thoroughly reviewed the last ten years question papers and then prepared the notes accordingly.

It can be tiresome for the students to keep each topic on their tips because of the substantial number of topics present in each chapter in the Class 11 Mathematics textbook. Hence, Trigonometric Functions Class 11 notes pdf plays a key role in making you memorize the topics of each chapter with ease. Download free Chapter 3 Maths Class 11 pdf with a single click on the pdf link given below.

A Few Glimpses of Class 11 Chapter 3 Trigonometric Functions

The word trigonometry is derived from the Greek word 'trignon' and 'metron' which implies 'measuring the slides of a triangle'. This topic was originally developed to solve the geometrical problem including triangles. Trigonometry was used by engineers, sea captains for navigation purposes, surveyors to find out new lands, etc. Presently, trigonometry is used in many areas such as for estimating the heights of tides in the ocean, designing electric current, the science of seismology, etc.

In your earlier classes, you must have studied trigonometric ratios of an acute angle as a ratio of the sides of the right-angle triangle. You must have also studied trigonometric identities and applications of trigonometric ratios.

In this chapter, you will extrapolate the concepts of trigonometry ratios to trigonometry functions and study their properties.

Trigonometric Functions

Trigonometry functions also known as circular functions and states the relationship between sides and angles of a right-angle triangle. It implies that the relationship between sides and angle of a right angle triangle is derived by these trigonometric functions. The angles of sine, cosine, and tangent are the primary classification of trigonometric functions. And, the other three functions such as cotangent, secant, and cosecant are derived from the primary trigonometric functions. There exists an inverse trigonometric function for each of the above-mentioned trigonometry functions.

Topic and Subtopics Covered in Class 11 Chapter 3 Trigonometric Functions

Let us know the different topics and subtopics covered in Class 11 Chapter 3 Trigonometric Functions.

3.1: Introduction to Chapter

3.2: Angles

3.2.1: Degree Measure

3.2.2: Radian Measure

3.2.3: Relation between radian and real numbers

3.2.4: Relation between degree and radian

3.3: Trigonometric Functions

3.3.1: Sign of Trigonometric Functions

3.3.2: Domain And Range of Trigonometric Functions

3.4: Trigonometric Functions of Sum and Difference of Two Angles

3.5: Trigonometric Equations

Download free Trigonometric Functions Class 11 Notes pdf now to get brief information on all the topics and subtopics covered in the chapter. With the help of these revision notes, you will understand all the above-discussed topics in a better way as these notes are properly arranged for easy preparation of the chapter. These revision notes are of utmost importance to refer while preparing for the exam. So download it now and minimize your stress.

Benefits of by Vedantu Class 11 Maths Notes Chapter 3

Some of the benefits of Class 11 Maths Notes Chapter 3 offered by Vedantu are discussed below:

Class 11 Maths Notes Chapter 3 provides an outline of the chapter in a short and precise manner.

Students can easily revise the important formulas and theorems related to the trigonometric function quickly and efficiently.

Students can save their valuable time by referring to the Mathematics notes for Class 11 Chapter 3.

Students can immediately recall all the important concepts of the chapter just by having a look at the revision notes.

Students can use the Class 11 Maths Notes Chapter 3 to revise the topic trigonometry rigorously just before the exam.

FAQs on Trigonometric Functions Class 11 Notes CBSE Maths Chapter 3 [Free PDF Download]

1. Name the Six Trigonometric Functions?

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

2. Name the Three Basic Trigonometric Functions?

The three basic trigonometric functions are Sine, Cosine, and Tangent.

3. What are the Different Formulas of Trigonometric Functions?

The different formulas of trigonometric functions are discussed below:

Sin λ = Opposite Side of angle λ / Hypotenuse Side of angle λ.

Cos λ = Adjacent Side of angle λ / Hypotenuse Side of angle λ.

Tan λ = Opposite Side of angle λ / Adjacent Side of angle λ.

Cot λ = Adjacent Side of angle λ / Opposite Side of angle λ.

Sec λ = Hypotenuse Side of angle λ / Adjacent Side of angle λ

Cosec λ = Hypotenuse Side of angle λ / Opposite Side of angle λ.

4. Calculate the Value of Sin λ, Cos λ, and Tan λ, if λ = 30 Degrees?

If the value of = 30 degrees, then,

Sin λ = Sin 30° = ½

Cos λ = Cos 30° = √3/2

Tan λ = Tan 30° = 1/√3

5. Where can I download the latest Chapter 3 Trigonometry notes of Class 11 Maths?

You can find Chapter 3 Trigonometry of Class 11 Maths Revision Notes online on Vedantu. This learning app focuses on helping students clear their concepts of Chapter 3 and also provides you with the revision notes in pdf format for them to study well and ace their final exams. To download the Class 11 Maths Chapter 3 notes, follow these steps -

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6. Can I download the Notes of  Chapter 3 Trigonometry of Class 11 Maths in PDF?

Yes, you can download the notes for Chapter 3 Trigonometry of Class 11 Maths as a PDF. Students don’t need to worry about writing the perfect answers in their exams as the revision notes prepared by Vedantu are just a click away. Vedantu provides good quality, informative and well-structured revision notes. You can sit at home and utilise all the benefits offered by Vedantu by downloading the notes for your reference.

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7. What are the topics in Chapter 3 Trigonometry of Class 11 Maths?

Chapter 3 Trigonometry of Class 11 Maths deals with the various trigonometric functions like sin, cos, tan and firstly gives a brief introduction of the chapter. The topics included in this chapter are the angles, the measure for radian and degree of the functions, the relation between the both, the sign of the trigonometric functions, their sum and differences, the equations relating to it and the relationship between real numbers and the radian values. All the topics and sub-topics are important for you to score good marks in exams.

8. What are trigonometric functions according to Chapter 3 Trigonometry of Class 11 Maths?

When you study Chapter 3 Trigonometry of Class 11 Maths, you will firstly have to keep clear concepts regarding the topics included in this chapter. So to begin with you will have to understand exactly the functions of trigonometry. These are used to determine the relationship between the angles of a triangle with the sides of that same right angled triangle. Another name for these trigonometric functions are circular functions.

9. What are the angles of trigonometry according to Chapter 3 Trigonometry of Class 11 Maths?

As you study Chapter 3 Trigonometry of Class 11 Maths, you will come across various new terms that are used in the trigonometry questions. Among these, the main thing is the angles of trigonometry. These are basically functions that help to relate triangle’s sides and angles. These are the sine, cosine and tangent, abbreviated as sin, cos, and tan. The reverse functions or angles are secant, cosecant and cotangent, abbreviated as sec, cosec, and cot.

CBSE Study Materials for Class 11

• Math Article

Trigonometric Functions Class 11

Trigonometric functions for class 11 notes are provided here. All the concepts in class 11 trigonometric functions are covered here as per the CBSE syllabus, which helps students while preparing for the examinations. Go through the topics given below and score good marks in the final examination. In this article, we have discussed topics like six important trigonometric functions, angle measures such as radians and degrees, sum and difference of two angles with many solved examples.

Trigonometric Functions Class 11 Topics

The topics and subtopics covered in trigonometric functions class 11 include:

• Introduction

Trigonometric Functions

• Trigonometric Functions of Sum and Difference of Two Angles

Introduction to Trigonometry

The word “ Trigonometry” is derived from the Greek word “trigon” and “metron”, which means measuring the sides of a triangle. Trigonometry is a branch of Mathematics which helps to study the relationship between the angles and side lengths of a right triangle. In the earlier classes, we have learnt the different trigonometric ratios of the right triangle. In trigonometric functions class 11, we are going to study the generalised concepts of trigonometric functions and their properties.

An angle is a measure which gives the rotation of a ray about its initial position. In other words, the measurement of an angle is the amount of rotation performed to obtain the terminal side from the initial side. The angle can be measured in two different ways, such as 

Degree: If rotation of the ray from the initial side to the terminal side is (1/360) th of the revolution, then the angle is considered to have a measure of one degree (1°).

Radians: The angle subtended at the centre by an arc length of one unit in a unit circle is considered to have a measure of 1 radian.

Relation Between Degree and Radian

As we know that the circle subtends at the centre of an angle whose radian measures 2π, and its degree measures 360 degrees. 

Hence, we get

2π = 360 degrees 

π radian = 180 degrees.

Therefore, 1 Radian = 180°/π 

Hence, 1 Radian = 57°16’, approximately

Also, 1 degree can be written as 

1° = π/180 radian = 0.01746 radians approximately.

The trigonometric functions are also called the angle functions, which relate the angles and the ratios of the sides of a right angle triangle . Considering the unit circle, the six important trigonometric functions are given as follows:

If “n” is an integer, then

Sin x =0 ⇒ x =n π

Cos x = 0 ⇒ x= (2n+1) π/2

The other trigonometric functions in terms of sine and cosine functions are given as follows:

  Tan x = sin x /cos x ⇒ x≠ (2n+1) π/2

Cosec x = 1/sin x ⇒ x ≠ nπ,

Sec x = 1/cos x ⇒  x≠ (2n+1) π/2

Cot x = cos x/sin x ⇒ x ≠ nπ

Sum and Difference of Two Angles

Some of the important expressions for the trigonometric functions of the sum and difference of two angles and related expressions are given as follows:

• Sin (-x) = – sin x
• Cos (-x) = cos x
• Cos (x+y) = cos x cos y – sin x sin y
• Cos (x-y) = cos x cos y + sin x sin y
• Cos (π/2 -x )= sin x
• Sin (π/2 -x ) = cos x
• Sin (x+y) = sin x cos y + cos x sin y
• Sin (x-y) = sin x cos y – cos x sin y
• Cos (π/2 + x ) = -sin x
• Sin (π/2 + x ) = cos x
• Cos (π-x) = -cos x 
• Sin (π-x) = sin x
• Cos (π+x) = -cos x
• Sin (π+x) = -sin x
• Cos (2π-x) = cos x
• Sin (2π-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 x + cot y)
• Cot (x-y) = (cot x cot y +1)/(cot y – cot x)
• Cos 2x = (1-tan 2 x)/(1+tan 2 x)
• Sin 2x = 2tanx/(1+tan 2 x)
• Tan 2x = 2tanx/(1-tan 2 x) 
• Sin 3x = 3 sin x – 4 sin 3 x
• Cos 3x =  4 cos 3 x – 3 cos x
• Tan 3x = (3 tanx -tan 3 x)/(1-3tan 2 x)

Video Lesson

Class 11 trigonometry.

Trigonometric Functions MCQs

Trigonometric Functions Class 11 Solved Examples

Go through the below given trigonometric functions class 11 solved problems:

Convert  6 radians into the degree measure.

Given : Radian = 6

We know that, π radian = 180°

Therefore 1 radian = 180°/π

Hence, 6 radians = (180°/π) × 6 degrees.

= (1080×7)/22 degrees

As we know, 1° = 60’

By simplifying the above equation, we get

 =343° 38′ 11″ approximately.

Thus, 6 radians is approximately equal to 343° 38′ 11″.

Example 2: 

Calculate the values of the other five trigonometric functions if cot x = -5/12, where x lies in the second quadrant.

Given: cot x = -5/12

From the above ratio, we can write tan x = -12/5

Sec 2 x = 1 + tan 2 x

= 1+ (12/5) 2

= 1+ (144/25)

Therefore, sec x = = ±13/5

Since, x lies in the second quadrant, the value of secant will be negative.

Hence, sec x = -13/5

Thus, cos x = -5/13

We know that tan x = sin x/cos x

Sin x = tan x cos x

Sin x = (-12/5) × (-5/13)

Sin x = 12/13

Thus, cosec x = 13/12

Therefore, the value of trigonometric functions are:

Sin x = 12/13, cos x = -5/13, tanx = -12/5, cosex x = 13/12, sec x = -13/5

Example 3: 

Calculate the value of cos (–1710°).

Given: cos (–1710°)

We know that the value of cos x repeats after the interval range of 2π; we can write:

cos (–1710°) =  cos [–1710° + (5 × 360°)]

cos (–1710°) = cos [–1710° + 1800°]

cos (–1710°) = cos 90° = 0

Therefore, the value of cos (–1710°) is 0.

Example 4: 

Prove that tan 3x tan 2x tan x = tan 3x – tan 2 x – tan x.

To prove: tan 3x tan 2x tan x = tan 3x – tan 2 x – tan x

We know that 3x can be written as 2x+x

Hence, tan 3x = tan (2x+x)

By using the trigonometric identity, the above expression is written as:

Tan 3x = (tan 2x + tan x)/(1-tan 2x tan x)

Now, cross multiply the above expression, and we get

Tan 3x – tan 3x tan 2x tan x = tan 2x + tan x

Simplify the above equation, we get

tan 3x tan 2x tan x = tan 3x – tan 2x – tan x.

Hence, tan 3x tan 2x tan x = tan 3x – tan 2x – tan x is proved.

Practice Problems

Solve the following class 11 trigonometric functions problems given below:

• Find the length of the minor arc of the chord, whose circle diameter is 40 cm and chord length is 20 cm.
• Determine the degree measure for the corresponding radian measure: (a)  -4   (b)7π/6
• Find the value of the other trigonometric functions if cos x = -⅗, where x lies in the third quadrant.
• Find the value of sin 31π/3.
• Prove that [sin (x+y)] /[sin (x-y)] = (tan x + tan y)/(tan x- tan y)

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