Class 11 Maths Chapter 3 Trigonometric Functions NCERT Solutions

Trigonometric Functions

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

= 300o

(iv) 7π/6

Here π radian = 180°

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:

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 radius r unit, if an arc of length l unit subtends an angle θ 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

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 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

By further calculation

= cosec 30o = 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 60o

= √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 75o

(ii) tan 15o

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 4x sin 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) = cot x (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. cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1

Solution:

23.

Solution:

Consider

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

By using the formula

24. cos 4x = 1 – 8sin2 x cos2 x

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 x – 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 x – 1

= 4 [(2 cos2 x – 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 cos6x – 48 cos4x + 18 cos2x – 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 + sin x) sin x + (cos 3x – cos x) cos x = 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:

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

8.

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: