Ncert Solutions for Class 11 Maths Pdf Chapter 3

NCERT Solutions For Class 11 Maths Chapter 3 Trigonometric Functions are available at BYJU'S, which are prepared by our expert teachers. All these solutions are written as per the latest update on term-wise CBSE Syllabus 2021-22 and its guidelines. BYJU'S provides step by step solutions by considering the different understanding levels of students and the marking scheme. Chapter 3 Trigonometric Functions comes under the second term portion of NCERT Class 11 Maths and is an important chapter for students. Though the chapter has more mathematical terms and formulae, BYJU'S made NCERT Solutions for Class 11 Maths easy for the students to understand and remember them, using tricks.

Trigonometry is developed to solve geometric problems involving triangles. Students cannot skip Trigonometric chapters since this is being used in many areas such as finding the heights of tides in the ocean, designing electronic circuits, etc., In the earlier classes, students have already studied trigonometric identities and applications of trigonometric ratios. In Class 11, trigonometric ratios are generalised to trigonometric function and their properties. However, these NCERT Solutions of BYJU'S help the students to attain more knowledge and score full marks in this chapter of term – II examination.

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

In this section, a few important terms are defined such as principal solution, general solution of trigonometric functions, which are explained using examples too.
Exercise 3.1 Solutions 7 Questions
Exercise 3.2 Solutions 10 Questions
Exercise 3.3 Solutions 25 Questions
Exercise 3.4 Solutions 9 Questions
Miscellaneous Exercise On Chapter 3 Solutions 10 Questions

Access NCERT Solutions for Class 11 Maths Chapter 3

Exercise 3.1 page: 54

1. Find the radian measures corresponding to the following degree measures:

(i) 25° (ii) – 47° 30′ (iii) 240° (iv) 520°

Solution:

(iv) 520°

2. Find the degree measures corresponding to the following radian measures (Use π = 22/7)

(i) 11/16

(ii) -4

(iii) 5π/3

(iv) 7π/6

Solution:

(i) 11/16

Here π radian = 180°

(ii) -4

Here π radian = 180°

(iii) 5π/3

Here π radian = 180°

We get

= 300^o

(iv) 7π/6

Here π radian = 180°

We get

= 210^o

3. A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?

Solution:

It is given that

No. of revolutions made by the wheel in

1 minute = 360

1 second = 360/60 = 6

We know that

The wheel turns an angle of 2π radian in one complete revolution.

In 6 complete revolutions, it will turn an angle of 6 × 2π radian = 12 π radian

Therefore, in one second, the wheel turns an angle of 12π radian.

4. Find the degree measure of the angle subtended at the centre of a circle of radius 100 cm by an arc of length 22 cm (Use π = 22/7).

Solution:

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.

Solution:

The dimensions of the circle are

Diameter = 40 cm

Radius = 40/2 = 20 cm

Consider AB be as the chord of the circle i.e. length = 20 cm

In ΔOAB,

Radius of circle = OA = OB = 20 cm

Similarly AB = 20 cm

Hence, ΔOAB is an equilateral triangle.

θ = 60° = π/3 radian

In a circle of radiusr unit, if an arc of lengthl unit subtends an angleθ radian at the centre

We get θ = 1/r

Therefore, the length of the minor arc of the chord is 20π/3 cm.

6. If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, find the ratio of their radii.

Solution:

7. Find the angle in radian though 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

Solution:

In a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then θ = 1/r

We know that r = 75 cm

(i) l = 10 cm

So we get

θ = 10/75 radian

By further simplification

θ = 2/15 radian

(ii) l = 15 cm

So we get

θ = 15/75 radian

By further simplification

θ = 1/5 radian

(iii) l = 21 cm

So we get

θ = 21/75 radian

By further simplification

θ = 7/25 radian

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Exercise 3.2 page: 63

Find the values of other five trigonometric functions in Exercises 1 to 5.

1. cos x = -1/2, x lies in third quadrant.

Solution:

2. sin x = 3/5, x lies in second quadrant.

Solution:

It is given that

sin x = 3/5

We can write it as

We know that

sin² x + cos² x = 1

We can write it as

cos² x = 1 – sin² x

3. cot x = 3/4, x lies in third quadrant.

Solution:

It is given that

cot x = 3/4

We can write it as

We know that

1 + tan² x = sec² x

We can write it as

1 + (4/3)² = sec² x

Substituting the values

1 + 16/9 = sec² x

cos² x = 25/9

sec x = ± 5/3

Here x lies in the third quadrant so the value of sec x will be negative

sec x = – 5/3

We can write it as

4. sec x = 13/5, x lies in fourth quadrant.

Solution:

It is given that

sec x = 13/5

We can write it as

We know that

sin² x + cos² x = 1

We can write it as

sin² x = 1 – cos² x

Substituting the values

sin² x = 1 – (5/13)²

sin² x = 1 – 25/169 = 144/169

sin² x = ± 12/13

Here x lies in the fourth quadrant so the value of sin x will be negative

sin x = – 12/13

We can write it as

5. tan x = -5/12, x lies in second quadrant.

Solution:

It is given that

tan x = – 5/12

We can write it as

We know that

1 + tan² x = sec² x

We can write it as

1 + (-5/12)² = sec² x

Substituting the values

1 + 25/144 = sec² x

sec² x = 169/144

sec x = ± 13/12

Here x lies in the second quadrant so the value of sec x will be negative

sec x = – 13/12

We can write it as

Find the values of the trigonometric functions in Exercises 6 to 10.

6. sin 765°

Solution:

We know that values of sin x repeat after an interval of 2π or 360°

So we get

By further calculation

= sin 45^o

= 1/ √ 2

7. cosec (–1410°)

Solution:

We know that values of cosec x repeat after an interval of 2π or 360°

So we get

By further calculation

= cosec 30^o = 2

8.

Solution:

We know that values of tan x repeat after an interval of π or 180°

So we get

By further calculation

We get

= tan 60^o

= √3

9.

Solution:

We know that values of sin x repeat after an interval of 2π or 360°

So we get

By further calculation

10.

Solution:

We know that values of tan x repeat after an interval of π or 180°

So we get

By further calculation

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Exercise 3.3 page: 73

Prove that:

1.

Solution:

2.

Solution:

Here

= 1/2 + 4/4

= 1/2 + 1

= 3/2

= RHS

3.

Solution:

4.

Solution:

5. Find the value of:

(i) sin 75^o

(ii) tan 15^o

Solution:

(ii) tan 15°

It can be written as

= tan (45° – 30°)

Using formula

Prove the following:

6.

Solution:

7.

Solution:

8.

Solution:

9.

Solution:

Consider

It can be written as

= sin x cos x (tan x + cot x)

So we get

10. sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x = cos x

Solution:

LHS = sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x

11.

Solution:

Consider

Using the formula

12. sin² 6x – sin² 4x = sin 2x sin 10x

Solution:

13. cos² 2x – cos² 6x = sin 4xsin 8x

Solution:

We get

= [2 cos 4x cos (-2x)] [-2 sin 4x sin (-2x)]

It can be written as

= [2 cos 4x cos 2x] [–2 sin 4x(–sin 2x)]

So we get

= (2 sin 4x cos 4x) (2 sin 2x cos 2x)

= sin 8x sin 4x

= RHS

14. sin 2x + 2sin 4x + sin 6x = 4cos² x sin 4x

Solution:

By further simplification

= 2 sin 4x cos (– 2x) + 2 sin 4x

It can be written as

= 2 sin 4x cos 2x + 2 sin 4x

Taking common terms

= 2 sin 4x (cos 2x + 1)

Using the formula

= 2 sin 4x (2 cos² x – 1 + 1)

We get

= 2 sin 4x (2 cos² x)

= 4cos² x sin 4x

= R.H.S.

15. cot 4x (sin 5x + sin 3x) = cotx (sin 5x – sin 3x)

Solution:

Consider

LHS = cot 4x (sin 5x + sin 3x)

It can be written as

Using the formula

= 2 cos 4x cos x

Hence, LHS = RHS.

16.

Solution:

Consider

Using the formula

17.

Solution:

18.

Solution:

19.

Solution:

20.

Solution:

21.

Solution:

22. cotx cot 2x – cot 2x cot 3x – cot 3x cotx = 1

Solution:

23.

Solution:

Consider

LHS = tan 4x = tan 2(2x)

By using the formula

24. cos 4x = 1 – 8sin² xcos² x

Solution:

Consider

LHS = cos 4x

We can write it as

= cos 2(2x)

Using the formula cos 2A = 1 – 2 sin²A

= 1 – 2 sin² 2x

Again by using the formula sin2A = 2sinA cos A

= 1 – 2(2 sin xcosx) ²

So we get

= 1 – 8 sin² x cos² x

= R.H.S.

25. cos 6x = 32 cos⁶x – 48 cos⁴x + 18 cos²x– 1

Solution:

Consider

L.H.S. = cos 6x

It can be written as

= cos 3(2x)

Using the formula cos 3A = 4 cos³A – 3 cos A

= 4 cos³ 2x – 3 cos2x

Again by using formula cos 2x = 2 cos²x– 1

= 4 [(2 cos²x– 1)³ – 3 (2 cos²x – 1)

By further simplification

= 4 [(2 cos²x) ³ – (1)³ – 3 (2 cos²x) ² + 3 (2 cos²x)] – 6cos²x + 3

We get

= 4 [8cos⁶ x – 1 – 12 cos⁴ x + 6 cos² x] – 6 cos² x + 3

By multiplication

= 32 cos⁶ x – 4 – 48 cos⁴ x + 24 cos²x – 6 cos² x + 3

On further calculation

= 32 cos⁶ x– 48 cos⁴ x + 18 cos² x – 1

= R.H.S.

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Exercise 3.4 PAGE: 78

Find the principal and general solutions of the following equations:

1. tan x = √3

Solution:

2. sec x = 2

Solution:

3. cot x = – √3

Solution:

4. cosec x = – 2

Solution:

Find the general solution for each of the following equations:

5. cos 4x = cos 2x

Solution:

6. cos 3x + cos x – cos 2x = 0

Solution:

7. sin 2x + cos x = 0

Solution:

It is given that

sin 2x + cos x = 0

We can write it as

2 sin x cos x + cos x = 0

cos x (2 sin x + 1) = 0

cos x = 0 or 2 sin x + 1 = 0

Let cos x = 0

8. sec² 2x = 1 – tan 2x

Solution:

It is given that

sec² 2x = 1 – tan 2x

We can write it as

1 + tan² 2x = 1 – tan 2x

tan² 2x + tan 2x = 0

Taking common terms

tan 2x (tan 2x + 1) = 0

Here

tan 2x = 0 or tan 2x + 1 = 0

If tan 2x = 0

tan 2x = tan 0

We get

2x = nπ + 0, where n ∈ Z

x = nπ/2, where n ∈ Z

tan 2x + 1 = 0

We can write it as

tan 2x = – 1

So we get

Here

2x = nπ + 3π/4, where n ∈ Z

x = nπ/2 + 3π/8, where n ∈ Z

Hence, the general solution is nπ/2 or nπ/2 + 3π/8, n ∈ Z.

9. sin x + sin 3x + sin 5x = 0

Solution:

It is given that

sin x + sin 3x + sin 5x = 0

We can write it as

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

Using the formula

By further calculation

2 sin 3x cos (-2x) + sin 3x = 0

It can be written as

2 sin 3x cos 2x + sin 3x = 0

By taking out the common terms

sin 3x (2 cos 2x + 1) = 0

Here

sin 3x = 0 or 2 cos 2x + 1 = 0

If sin 3x = 0

3x = nπ, where n ∈ Z

We get

x = nπ/3, where n ∈ Z

If 2 cos 2x + 1 = 0

cos 2x = – 1/2

By further simplification

= – cos π/3

= cos (π – π/3)

So we get

cos 2x = cos 2π/3

Here

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Miscellaneous Exercise page: 81

Prove that:

1.

Solution:

We get

= 0

= RHS

2. (sin 3x+ sinx) sinx+ (cos 3x– cosx) cosx= 0

Solution:

Consider

LHS = (sin 3x + sin x) sin x + (cos 3x – cos x) cos x

By further calculation

= sin 3x sin x + sin² x + cos 3x cos x – cos² x

Taking out the common terms

= cos 3x cos x + sin 3x sin x – (cos² x – sin² x)

Using the formula

cos (A – B) = cos A cos B + sin A sin B

= cos (3x – x) – cos 2x

So we get

= cos 2x – cos 2x

= 0

= RHS

3.

Solution:

Consider

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

By expanding using formula we get

= cos² x + cos² y + 2 cos x cos y + sin² x + sin² y – 2 sin x sin y

Grouping the terms

= (cos² x + sin² x) + (cos² y + sin² y) + 2 (cos x cos y – sin x sin y)

Using the formula cos (A + B) = (cos A cos B – sin A sin B)

= 1 + 1 + 2 cos (x + y)

By further calculation

= 2 + 2 cos (x + y)

Taking 2 as common

= 2 [1 + cos (x + y)]

From the formula cos 2A = 2 cos² A – 1

4.

Solution:

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

By expanding using formula

= cos² x + cos² y – 2 cos x cos y + sin² x + sin² y – 2 sin x sin y

Grouping the terms

= (cos² x + sin² x) + (cos² y + sin² y) – 2 (cos x cos y + sin x sin y)

Using the formula cos (A – B) = cos A cos B + sin A sin B

= 1 + 1 – 2 [cos (x – y)]

By further calculation

= 2 [1 – cos (x – y)]

From formula cos 2A = 1 – 2 sin² A

5. sin x + sin 3x + sin 5x + sin 7x = 4 cos x cos 2x sin 4x

Solution:

6.

Solution:

7.

Solution:

8. Find sin x/2, cos x/2 and tan x/2 in each of the following:

Solution:

cos x = -3/5

From the formula

9. cos x = -1/3, x in quadrant III

Solution:

10. sin x = 1/4, x in quadrant II

Solution:

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NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions

This chapter has 6 exercises and a miscellaneous exercise to help students understand the concepts related to Trigonometric Functions clearly. BYJU'S provides all the concepts and solutions for Chapter 3 of Class 11 Maths with clear explanations and formulas. The PDF of Maths  NCERT Solutions for Class 11 Chapter 3 includes the topics and sub-topics listed below.
3.1 Introduction
The basic trigonometric ratios and identities are given here along with the applications of trigonometric ratios in solving the word problems related to heights and distances.
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
In this section, different terms related to trigonometry are discussed such as terminal side, initial sides, measuring an angle in degrees and radian, etc.
3.3 Trigonometric Functions
3.3.1 Sign of trigonometric functions
3.3.2 Domain and range of trigonometric functions
After studying this section, students are able to understand the generalised trigonometric functions with signs. Also, they can gain knowledge on domain and range of trigonometric functions with examples.
3.4 Trigonometric Functions of Sum and Difference of Two Angles
This section contains formulas related to the sum and difference of two angles in trigonometric functions.

Key Features of NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions

Studying the Trigonometric Functions of Class 11 enables the students to understand the following:

• Introduction to Trigonometric Functions
• Positive and negative angles
• Measuring angles in radians and in degrees and conversion of one into other
• Definition of trigonometric functions with the help of unit circle
• Truth of the sin 2x + cos 2x = 1, for all x
• Signs of trigonometric functions
• Domain and range of trigonometric functions
• Graphs of Trigonometric Functions
• Expressing sin (x±y) and cos (x±y) in terms of sin x, sin y, cos x & cosy and their simple application, identities related to sin 2x, cos 2x, tan 2x, sin 3x, cos 3x and tan 3x
• The general solution of trigonometric equations of the type sin y = sin a, cos y = cos a and tan y = tan a

Frequently Asked Questions on NCERT Solutions for Class 11 Maths Chapter 3

What are the topics discussed in Chapter 3 of NCERT Solutions for Class 11 Maths as per the latest update of the second term CBSE Syllabus?

The topics discussed in Chapter 3 of NCERT Solutions for Class 11 Maths as per the latest term-wise CBSE Syllabus (2021-22) are:
1. Introduction
2. Angles
3. Trigonometric Functions
4. Trigonometric Functions of sum and difference of two angles

How many exercises are there in Chapter 3 of NCERT Solutions for Class 11 Maths?

There are 4 exercises and one miscellaneous exercise in Chapter 3 of NCERT Solutions for Class 11 Maths. The number of questions in each exercise are mentioned below.
Exercise 3.1 – 7 Questions
Exercise 3.2 – 10 Questions
Exercise 3.3 – 25 Questions
Exercise 3.4 – 9 Questions
Miscellaneous Exercise – 10 Questions

What will I learn in Chapter 3 Trigonometric Functions of NCERT Solutions for Class 11 Maths?

Chapter 3 Trigonometric Functions of NCERT Solutions for Class 11 Maths covers the complex topics of trigonometric functions and their uses. This chapter consists of 4 sub sections which deals with topics like measuring angles in radians and degrees and their interconversion. These concepts are explained in a detailed manner to improve logical thinking abilities among students.

Ncert Solutions for Class 11 Maths Pdf Chapter 3

Source: https://byjus.com/ncert-solutions-class-11-maths/chapter-3-trigonometric-functions/

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