NCERT Solution For Class 11 Maths Chapter 3 Trigonometric Functions

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 guidelines of CBSE. BYJU’S provides step by step solutions by considering the different understanding levels of students and the marking scheme. Chapter 3 Trigonometric Functions of NCERT Class 11 Maths is an important chapter for students. Though the chapter has more mathematical terms and formulae, BYJU’S made NCERT Solutions for Class 11 Maths is 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 from this chapter.

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

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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:

NCERT Solutions for Class 11 Chapter 3 Ex 3.1 Image 1

NCERT Solutions for Class 11 Chapter 3 Ex 3.1 Image 2

NCERT Solutions for Class 11 Chapter 3 Ex 3.1 Image 3

(iv) 520°

NCERT Solutions for Class 11 Chapter 3 Ex 3.1 Image 4

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°

NCERT Solutions for Class 11 Chapter 3 Ex 3.1 Image 5

(ii) -4

Here π radian = 180°

NCERT Solutions for Class 11 Chapter 3 Ex 3.1 Image 6

(iii) 5π/3

Here π radian = 180°

NCERT Solutions for Class 11 Chapter 3 Ex 3.1 Image 7

We get

= 300o

(iv) 7π/6

Here π radian = 180°

NCERT Solutions for Class 11 Chapter 3 Ex 3.1 Image 8

We get

= 210o

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 = 360/6 = 60

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:

NCERT Solutions for Class 11 Chapter 3 Ex 3.1 Image 9

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

NCERT Solutions for Class 11 Chapter 3 Ex 3.1 Image 10

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 radius r unit, if an arc of length l unit subtends an angle θ radian at the centre

We get θ = 1/r

NCERT Solutions for Class 11 Chapter 3 Ex 3.1 Image 11

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:

NCERT Solutions for Class 11 Chapter 3 Ex 3.1 Image 12

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


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:

NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 1

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

Solution:

It is given that

sin x = 3/5

We can write it as

NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 2

We know that

sin2 x + cos2 x = 1

We can write it as

cos2 x = 1 – sin2 x

NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 3

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

Solution:

It is given that

cot x = 3/4

We can write it as

NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 4

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

NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 5

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

Solution:

It is given that

sec x = 13/5

We can write it as

NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 6

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

NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 7

NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 8

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

Solution:

It is given that

tan x = – 5/12

We can write it as

NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 9

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

NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 10

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

NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 11

By further calculation

= sin 45o

= 1/  2

7. cosec (–1410°)

Solution:

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

So we get

NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 12

By further calculation

NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 13

= cosec 30o = 2

8. NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 14

Solution:

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

So we get

NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 15

By further calculation

NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 16

We get

= tan 60o

= 3

9. NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 17

Solution:

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

So we get

NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 18

By further calculation

NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 19

10. NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 20

Solution:

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

So we get

NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 21

By further calculation

NCERT Solutions for Class 11 Chapter 3 Ex 3.2 Image 22


Exercise 3.3 page: 73

Prove that:

1.

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 1

Solution:

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 2

2.

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 3

Solution:

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 4

Here

= 1/2 + 4/4

= 1/2 + 1

= 3/2

= RHS

3.

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 5

Solution:

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 6

4.

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 7

Solution:

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 8

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 9

5. Find the value of:

(i) sin 75o

(ii) tan 15o

Solution:

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 10

(ii) tan 15°

It can be written as

= tan (45° – 30°)

Using formula

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 11

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 12

Prove the following:

6.

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 13

Solution:

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 14

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 15

7.

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 16

Solution:

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 17

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 18

8.

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 19

Solution:

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 20

9.

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 21

Solution:

Consider

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 22

It can be written as

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

So we get

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 23

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 

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 24

11.

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 25

Solution:

Consider

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 26

Using the formula

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 27

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

Solution:

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 28

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 29

13. cos2 2x – cos2 6x = sin 4sin 8x

Solution:

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 30

We get

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

It can be written as

= [2 cos 4x cos 2x] [–2 sin 4(–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:

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 31

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) = cot x (sin 5x – sin 3x)

Solution:

Consider

LHS = cot 4x (sin 5x + sin 3x)

It can be written as

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 32

Using the formula

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 33

= 2 cos 4x cos x

Hence, LHS = RHS.

16.

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 34

Solution:

Consider

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 35

Using the formula

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 36

17.

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 37

Solution:

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 28

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 39

18.

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 40

Solution:

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 41

19.

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 42

Solution:

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 43

20.

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 44

Solution:

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 45

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 46

21.

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 47

Solution:

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 48

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 49

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

Solution:

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 50

23.

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 51

Solution:

Consider

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

By using the formula

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 52

NCERT Solutions for Class 11 Chapter 3 Ex 3.3 Image 53

24. cos 4x = 1 – 8sincosx

Solution:

Consider

LHS = cos 4x

We can write it as

= cos 2(2x)

Using the formula cos 2A = 1 – 2 sin2 A

= 1 – 2 sin2 2x 

Again by using the formula sin2A = 2sin A cos A

= 1 – 2(2 sin x cos x) 2 

So we get

= 1 – 8 sin2x cos2x

= R.H.S.

25. cos 6x = 32 cos6 x – 48 cos4 x + 18 cos2 – 1

Solution:

Consider

L.H.S. = cos 6x

It can be written as

= cos 3(2x)

Using the formula cos 3A = 4 cos3 A – 3 cos A

= 4 cos3 2x – 3 cos 2x 

Again by using formula cos 2x = 2 cos2 – 1

= 4 [(2 cos2 – 1)3 – 3 (2 cos2 x – 1)

By further simplification

= 4 [(2 cos2 x) 3 – (1)3 – 3 (2 cos2 x) 2 + 3 (2 cos2 x)] – 6cos2 x + 3

We get

= 4 [8cos6x – 1 – 12 cos4x + 6 cos2x] – 6 cos2x + 3

By multiplication

= 32 cos6x – 4 – 48 cos4x + 24 cos2 x – 6 cos2x + 3

On further calculation

= 32 cos6– 48 cos4x + 18 cos2x – 1

= R.H.S.


Exercise 3.4 PAGE: 78

Find the principal and general solutions of the following equations:

1. tan x = √3

Solution:

NCERT Solutions for Class 11 Chapter 3 Ex 3.4 Image 1

2. sec x = 2

Solution:

NCERT Solutions for Class 11 Chapter 3 Ex 3.4 Image 2

NCERT Solutions for Class 11 Chapter 3 Ex 3.4 Image 3

3. cot x = – √3

Solution:

NCERT Solutions for Class 11 Chapter 3 Ex 3.4 Image 4

NCERT Solutions for Class 11 Chapter 3 Ex 3.4 Image 5

4. cosec x = – 2

Solution:

NCERT Solutions for Class 11 Chapter 3 Ex 3.4 Image 6

NCERT Solutions for Class 11 Chapter 3 Ex 3.4 Image 7

Find the general solution for each of the following equations:

5. cos 4x = cos 2x

Solution:

NCERT Solutions for Class 11 Chapter 3 Ex 3.4 Image 8

6. cos 3x + cos x – cos 2x = 0

Solution:

NCERT Solutions for Class 11 Chapter 3 Ex 3.4 Image 9

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

NCERT Solutions for Class 11 Chapter 3 Ex 3.4 Image 10

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

NCERT Solutions for Class 11 Chapter 3 Ex 3.4 Image 11

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

NCERT Solutions for Class 11 Chapter 3 Ex 3.4 Image 12

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

NCERT Solutions for Class 11 Chapter 3 Ex 3.4 Image 13


Miscellaneous Exercise page: 81

Prove that:

1.

NCERT Solutions for Class 11 Chapter 3 Miscellaneous Ex Image 1

Solution:

NCERT Solutions for Class 11 Chapter 3 Miscellaneous Ex Image 2

NCERT Solutions for Class 11 Chapter 3 Miscellaneous Ex Image 3

We get

NCERT Solutions for Class 11 Chapter 3 Miscellaneous Ex Image 4

= 0

= RHS

2. (sin 3+ sin x) sin + (cos 3– cos x) cos = 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.

NCERT Solutions for Class 11 Chapter 3 Miscellaneous Ex Image 5

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

NCERT Solutions for Class 11 Chapter 3 Miscellaneous Ex Image 6

4.

NCERT Solutions for Class 11 Chapter 3 Miscellaneous Ex Image 7

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

NCERT Solutions for Class 11 Chapter 3 Miscellaneous Ex Image 8

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

Solution:

NCERT Solutions for Class 11 Chapter 3 Miscellaneous Ex Image 9

NCERT Solutions for Class 11 Chapter 3 Miscellaneous Ex Image 10

6.

NCERT Solutions for Class 11 Chapter 3 Miscellaneous Ex Image 11

Solution:

NCERT Solutions for Class 11 Chapter 3 Miscellaneous Ex Image 12

7.

NCERT Solutions for Class 11 Chapter 3 Miscellaneous Ex Image 13

Solution:

NCERT Solutions for Class 11 Chapter 3 Miscellaneous Ex Image 14

NCERT Solutions for Class 11 Chapter 3 Miscellaneous Ex Image 15

NCERT Solutions for Class 11 Chapter 3 Miscellaneous Ex Image 16

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

8.

NCERT Solutions for Class 11 Chapter 3 Miscellaneous Ex Image 17

Solution:

NCERT Solutions for Class 11 Chapter 3 Miscellaneous Ex Image 18

cos x = -3/5

From the formula

NCERT Solutions for Class 11 Chapter 3 Miscellaneous Ex Image 19

NCERT Solutions for Class 11 Chapter 3 Miscellaneous Ex Image 20

NCERT Solutions for Class 11 Chapter 3 Miscellaneous Ex Image 21

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

Solution:

NCERT Solutions for Class 11 Chapter 3 Miscellaneous Ex Image 22

NCERT Solutions for Class 11 Chapter 3 Miscellaneous Ex Image 23

NCERT Solutions for Class 11 Chapter 3 Miscellaneous Ex Image 24

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

Solution:

NCERT Solutions for Class 11 Chapter 3 Miscellaneous Ex Image 25

NCERT Solutions for Class 11 Chapter 3 Miscellaneous Ex Image 26

NCERT Solutions for Class 11 Chapter 3 Miscellaneous Ex Image 27

NCERT Solutions for Class 11 Chapter 3 Miscellaneous Ex Image 28

NCERT Solutions for Class 11 Chapter 3 Miscellaneous Ex Image 29


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

1 Comment

  1. It so best for us it shows shortest answer thank you for the help to solve every problem

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