**Exercise 3.1 : Solutions of Questions on Page Number : 54**

**Q1 :Find the radian measures corresponding to the following degree measures:**

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

**Answer :
(i)** 25°

We know that 180° = π radian

**(ii)**-47° 30′

-47° 30′ = degree [1° = 60′]

degree

Since 180° = π radian

**(iii)** 240°

We know that 180° = π radian

**(iv)** 520°

We know that 180° = π radian

**Q2 :Find the degree measures corresponding to the following radian measures**

**(i) (ii) – 4
(iii) (iv)**

**Answer**:

**(i)**We know that π radian = 180°

**(ii)**– 4

We know that π radian = 180°

**(iii)**

We know that π radian = 180°

**(iv)**

We know that π radian = 180°

**Q3 :A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?**

**Answer :**

Number of revolutions made by the wheel in 1 minute = 360

∴Number of revolutions made by the wheel in 1 second =

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.,

12 π radian

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

**Q4 :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.**

**Answer :
**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

Therefore, forr = 100 cm, l = 22 cm, we have

Thus, the required angle is 12°36“².

**Q5 :In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of minor arc of the chord.**

**Answer :
**Diameter of the circle = 40 cm

∴Radius (r) of the circle =

Let AB be a chord (length = 20 cm) of the circle.

In ΔOAB, OA = OB = Radius of circle = 20 cm

Also, AB = 20 cm

Thus, ΔOAB is an equilateral triangle.

∴θ = 60° =

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

Thus, the length of the minor arc of the chord is

**Q6 :If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, find the ratio of their radii.**

**Answer :
**Let the radii of the two circles be and. 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, 60° = and 75° =

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

Thus, the ratio of the radii is 5:4.

**Q7 :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**

**Answer :**

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

It is given that r = 75 cm

**(i)** Here, l = 10 cm

**(ii)** Here, l = 15 cm

**(iii)** Here, l = 21 cm

**Exercise 3.2 : Solutions of Questions on Page Number : 63**

** Q1 :Find the values of other five trigonometric functions if , x lies in third quadrant.**

**Answer :
**Since x lies in the 3rd quadrant, the value of sin x will be negative.

**Q2 :Find the values of other five trigonometric functions if , x lies in second quadrant.**

**Answer :
**Since x lies in the 2nd quadrant, the value of cos x will be negative

**Q3 :Find the values of other five trigonometric functions if , x lies in third quadrant.**

**Answer :
**Since x lies in the 3rd quadrant, the value of sec x will be negative.

**Q4 :Find the values of other five trigonometric functions if , x lies in fourth quadrant.**

**Answer :
**Since x lies in the 4th quadrant, the value of sin x will be negative.

**Q5 :Find the values of other five trigonometric functions if , x lies in second quadrant.**

**Answer** :

Since x lies in the 2nd quadrant, the value of sec x will be negative.

∴sec x =

**Q6 :Find the value of the trigonometric function sin 765°**

**Answer :**

It is known that the values of sin x repeat after an interval of 2π or 360°.

**Q7 :Find the value of the trigonometric function cosec (-1410°)**

**Answer** :

It is known that the values of cosec x repeat after an interval of 2π or 360°.

**Q8 :Find the value of the trigonometric function **

**Answer :**

It is known that the values of tan x repeat after an interval of π or 180°.

**Q9 :Find the value of the trigonometric function **

**Answer :**

It is known that the values of sin x repeat after an interval of 2π or 360°.

**Q10 :Find the value of the trigonometric function **

**Answer :
**It is known that the values of cot x repeat after an interval of π or 180°.

**Exercise 3.3 : Solutions of Questions on Page Number : 73**

** Q1 : **

**Answer** :

L.H.S. =

**Q2 :Prove that **

**Answer :**

L.H.S. =

**Q3 :Prove that **

**Answer :**

L.H.S. =

**Q4 :Prove that **

**Answer :**

L.H.S =

**Q5 :Find the value of:**

**(i) sin 75°**

**(ii) tan 15°**

**Answer :**

**(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]

**(ii)** tan 15° = tan (45° – 30°)

**Q6 :Prove that: **

**Answer :
**

**Q7 :Prove that: **

**Answer :
**It is known that

∴L.H.S. =

**Q8 :Prove that **

**Answer : **

**Q9 : **

**Answer :**

L.H.S. =

**Q10 :Prove that sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x = cos x**

**Answer :
**L.H.S. = sin (n + 1)x sin(n + 2)x + cos (n + 1)x cos(n + 2)x

4

**Q11 :Prove that **

**Answer :**

It is known that

∴L.H.S. =

**Q12 :Prove that sin ^{2} 6x – sin_{2} 4x = sin 2x sin 10x**

**Answer :**

It is known that

∴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.

**Q13 :Prove that cos ^{2} 2x – cos^{2} 6x = sin 4x sin 8x**

**Answer :**

It is known that

∴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 sin 4x cos 4x) (2 sin 2x cos 2x)

= sin 8x sin 4x

= R.H.S.

**Q14 :Prove that sin 2x + 2sin 4x + sin 6x = 4cos ^{2} x sin 4x**

**Answer :**

L.H.S. = sin 2x + 2 sin 4x + sin 6x

= [sin 2x + sin 6x] + 2 sin 4x

= 2 sin 4x cos (- 2x) + 2 sin 4x

= 2 sin 4x cos 2x + 2 sin 4x

= 2 sin 4x (cos 2x + 1)

= 2 sin 4x (2 cos

^{2}x – 1 + 1)

= 2 sin 4x (2 cos

^{2}x)

= 4cos

^{2}x sin 4x

= R.H.S.

**Q15 :Prove that cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)**

**Answer :**

L.H.S = cot 4x (sin 5x + sin 3x)

= 2 cos 4x cos x

R.H.S. = cot x (sin 5x – sin 3x)

= 2 cos 4x. cos x

L.H.S. = R.H.S.

**Q16 :Prove that **

**Answer :
**It is known that

∴L.H.S =

**Q17 :Prove that **

**Answer :
**It is known that

∴L.H.S. =

**Q18 :Prove that **

**Answer :**

It is known that

∴L.H.S. =

**Q19 :Prove that **

**Answer :
**It is known that

∴L.H.S. =

**Q20 :Prove that **

**Answer :
**It is known that

∴L.H.S. =

**Q21 :Prove that **

**Answer :
**L.H.S. =

**Q22 :Prove that cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1**

**Answer :
**L.H.S. = cot x cot 2x – cot 2x cot 3x – cot 3x cot x

= cot x cot 2x – cot 3x (cot 2x + cot x)

= cot x cot 2x – cot (2x + x) (cot 2x + cot x)

= cot x cot 2x – (cot 2x cot x – 1)

= 1 = R.H.S.

**Q23 :Prove that **

**Answer :
**It is known that

∴L.H.S. = tan 4x = tan 2(2x)

**Q24 :Prove that cos 4x = 1 – 8sin ^{2} x cos^{2} x**

**Answer :**

L.H.S. = cos 4x

= cos 2(2x)

= 1 – 2 sin

^{2}2x [cos 2A = 1 – 2 sin

^{2}A]

= 1 – 2(2 sin x cos x)

^{2}[sin2A = 2sin A cosA]

= 1 – 8 sin

^{2}x cos

^{2}x

= R.H.S.

**Q25 :Prove that: cos 6x = 32 cos ^{6} x – 48 cos^{4} x + 18 cos^{2} x – 1**

**Answer :**

L.H.S. = cos

^{6}x

= cos 3(2x)

= 4 cos

^{3}2x – 3 cos 2x [cos 3A = 4 cos

^{3}A – 3 cos A]

= 4 [(2 cos

^{2}x – 1)

^{3}– 3 (2 cos

^{2}x – 1) [cos 2x = 2 cos

^{2}x – 1]

= 4 [(2 cos

^{2}x)

^{3}– (1)

^{3}– 3 (2 cos

^{2}x)

^{2}+ 3 (2 cos

^{2}x)] – 6cos

^{2}x + 3

= 4 [8cos

^{6}x – 1 – 12 cos

^{4}x + 6 cos

^{2}x] – 6 cos

^{2}x + 3

= 32 cos

^{6}x – 4 – 48 cos

^{4}x + 24 cos

^{2}x – 6 cos

^{2}x + 3

= 32 cos

^{6}x – 48 cos

^{4}x + 18 cos

^{2}x – 1

= R.H.S

**Exercise 3.4 : Solutions of Questions on Page Number : 78**

** Q1 :Find the principal and general solutions of the equation **

**Answer :
**Therefore, the principal solutions are x = and

Therefore, the general solution is

**Q2 :Find the principal and general solutions of the equation **

**Answer :
**Therefore, the principal solutions are x = and

Therefore, the general solution is, where n ∈ Z

**Q3 :Find the principal and general solutions of the equation **

**Answer :
**Therefore, the principal solutions are x = and

Therefore, the general solution is

**Q4 :Find the general solution of cosec x = -2**

**Answer :
**cosec x= -2

Therefore, the principal solutions are x =

Therefore, the general solution is

**Q5 :Find the general solution of the equation cos 4x = cos 2x**

**Answer :
**

**Q6 :Find the general solution of the equation cos 3x + cos x – cos 2x = 0**

**Answer :
**cos 3x + cos x – cos 2x = 0

**Q7 :Find the general solution of the equation sin 2x + cos x = 0**

**Answer :
**sin 2x + cos x = 0

Therefore, the general solution is

**Q8 :Find the general solution of the equation sec ^{2 }2x = 1 – tan 2x
**

**Answer :**

sec2 2x = 1 – tan 2x

**Therefore, the general solution is**

**Q9 :Find the general solution of the equation sin x + sin 3x + sin 5x =0**

**Answer :
**sin x + sin 3x + sin 5x =0

**Therefore, the general solution is**

**Exercise Miscellaneous : Solutions of Questions on Page Number : 81**

** Q1 :Prove that: **

**Answer :**

L.H.S.

= 0 = R.H.S

**Q2 :Prove that: (sin 3x + sin x) sin x + (cos 3x – cos x) cos x = 0**

**Answer :
**L.H.S.= (sin 3x + sin x) sin x + (cos 3x – cos x) cos x

= RH.S.

**Q3 :Prove that: **

**Answer :**

L.H.S. =

**Q4 :Prove that: **

**Answer :**

L.H.S. =

**Q5 :Prove that: **

**Answer :**

It is known that

∴L.H.S. =

**Q6 :Prove that: **

**Answer :**

It is known that

L.H.S. =

= tan 6x

= R.H.S.

**Q7 :Prove that: **

**Answer :**

L.H.S. =

**Q8 : , x in quadrant II**

**Answer :**

Here, x is in quadrant II.

i.e.,

Therefore, are all positive.

As x is in quadrant II, cosx is negative.

∴

Thus, the respective values of are

**Q9 :Find for , x in quadrant III**

**Answer :**

Here, x is in quadrant III.

Therefore, and are negative, where as is positive.

Now,

Thus, the respective values of are

**Q10 :Find for , x in quadrant II**

**Answer :**

Here, x is in quadrant II.

Therefore,, and are all positive.

[cosx is negative in quadrant II]

Thus, the respective values of are