Exercise 3.1

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 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

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

Exercise 3.3

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 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.

Exercise 3.4

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