3 Trigonometric Funtions

Exercise 3.1

Question 1:

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

Question 2:

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°

Question 3:

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.

Question 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.

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′.

Question 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.

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.

Question 6:

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 r₁, while let an arc of length l subtend an angle of 75° at the centre of the circle of radius r₂.

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.

Question 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

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

Question 1:

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

Answer :

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

Question 2:

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

Answer :

Since x lies in the 2^nd quadrant, the value of cos x will be negative

Question 3:

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

Answer :

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

Question 4:

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

Answer :

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

Question 5:

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

Answer :

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

∴sec x =

Question 6:

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°.

Question 7:

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°.

Question 8:

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°.

Question 9:

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°.

Question 10:

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

Question 1:

Answer :

L.H.S. = 

Question 2:

Prove that 

Answer :

L.H.S. = 

Question 3:

Prove that 

Answer :

L.H.S. =

Question 4:

Prove that 

Answer :

L.H.S =

Question 5:

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°)

Question 6:

Prove that:

Answer :

Question 7:

Prove that: 

Answer :

It is known that 

∴L.H.S. =

Question 8:

Prove that

Answer :

Question 9:

Answer :

L.H.S. = 

Question 10:

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

Question 11:

Prove that

Answer :

It is known that.

∴L.H.S. = 

Question 12:

Prove that sin² 6x – sin² 4x = sin 2x sin 10x

Answer :

It is known that 

∴L.H.S. = sin²6x – sin²4x

= (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.

Question 13:

Prove that cos² 2x – cos² 6x = sin 4x sin 8x

Answer :

It is known that 

∴L.H.S. = cos² 2x – cos² 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.

Question 14:

Prove that sin 2x + 2sin 4x + sin 6x = 4cos² 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 cos² x – 1 + 1)

= 2 sin 4x (2 cos² x)

= 4cos² x sin 4x

= R.H.S.

Question 15:

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.

Question 16:

Prove that 

Answer :

It is known that

∴L.H.S =

Question 17:

Prove that 

Answer :

It is known that

∴L.H.S. =

Question 18:

Prove that 

Answer :

It is known that

∴L.H.S. =

Question 19:

Prove that 

Answer :

It is known that

∴L.H.S. =

Question 20:

Prove that 

Answer :

It is known that

∴L.H.S. = 

Question 21:

Prove that 

Answer :

L.H.S. =

Question 22:

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.

Question 23:

Prove that 

Answer :

It is known that.

∴L.H.S. = tan 4x = tan 2(2x)

Question 24:

Prove that cos 4x = 1 – 8sin² x cos² x

Answer :

L.H.S. = cos 4x

= cos 2(2x)

= 1 – 2 sin² 2x [cos 2A = 1 – 2 sin² A]

= 1 – 2(2 sin x cos x)² [sin2A = 2sin A cosA]

= 1 – 8 sin²x cos²x

= R.H.S.

Question 25:

Prove that: cos 6x = 32 cos⁶ x – 48 cos⁴ x + 18 cos² x – 1

Answer :

L.H.S. = cos 6x

= cos 3(2x)

= 4 cos³ 2x – 3 cos 2x [cos 3A = 4 cos³ A – 3 cos A]

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

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

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

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

= 32 cos⁶x – 48 cos⁴x + 18 cos²x – 1

= R.H.S.

Exercise 3.4

Question 1:

Find the principal and general solutions of the equation 

Answer :

Therefore, the principal solutions are x =and.

Therefore, the general solution is

Question 2:

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

Question 3:

Find the principal and general solutions of the equation 

Answer :

Therefore, the principal solutions are x = and.

Therefore, the general solution is 

Question 4:

Find the general solution of cosec x = –2

Answer :

_{cosec x = –2}

Therefore, the principal solutions are x =.

Therefore, the general solution is

Question 5:

Find the general solution of the equation 

Answer :

Question 6:

Find the general solution of the equation 

Answer :

Question 7:

Find the general solution of the equation 

Answer :

Therefore, the general solution is.

Question 8:

Find the general solution of the equation 

Answer :

Therefore, the general solution is.

Question 9:

Find the general solution of the equation 

Answer :

Therefore, the general solution is

Miscellaneous

Question 1:

Prove that: 

Answer :

L.H.S.

= 0 = R.H.S

Question 2:

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.

Question 3:

Prove that: 

Answer :

L.H.S. = 

Question 4:

Prove that: 

Answer :

L.H.S. = 

Question 5:

Prove that: 

Answer :

It is known that.

∴L.H.S. = 

Question 6:

Prove that: 

Answer :

It is known that

.

L.H.S. = 

= tan 6x

= R.H.S.