# Class 11 Maths NCERT Solutions for Chapter 3 Trigonometric Functions Miscellaneous Exercise

### Trigonometric Functions Miscellaneous Exercise Solutions

1. Prove that : 2 cos(Ï€/3) cos(9Ï€/13) + cos(3Ï€/13) + cos (5Ï€/13) = 0

Solution

L.H.S

= 0 = R.H.S

2. Prove that: (sin 3+ sin x) sin + (cos 3– cos x) cos = 0
Solution
L.H.S
= (sin 3x + sin x)sin x + (cos 3x - cos x) cos x
= sin 3x sin x + sin2x + cos3x cosx – cos2
= cos 3x cos x + sin 3x sin x - (cos2x - sin2x)
= cos(3x - x) - cos2x  [cos (A - B) = cosA cosB + sin A sinB]
= cos 2x - cos 2x
= 0
R.H.S

3. Prove that (cos x + cos y)2 + (sin x - sin y)2 = 4 cos2[(x+y)/2]
Solution
L.H.S = (cos x + cos y)2 + (sin x - sin y)2
= cos2x + cos2y + 2 cosx cosy + sin2x + sin2y - 2sin x siny
= (cos2x + sin2x) + (cos2y + sin2y) + 2(cos x cos y - sin x sin y)
= 1 + 1 + 2 cos(x + y)  [cos(A + B) = cos A cos B - sin A sin B)]
= 2 + 2cos (x + y)
= 2[1 + cos (x + y)]
= 2[1 + 2 cos2{(x+y)/2}- 1]  [cos2A = 2 cos2A - 1]
= 4 cos2{(x + y)/2} = R.H.S

4. Prove that : (cos x - cos y)2 + (sin x - sin y)2  = 4 sin2{(x - y)/2}
Solution
L.H.S = (cos x - cos y)2  + (sin x - sin y)2
= cos2x + cos2y - 2 cos x cos y + sin2x + sin2y - 2sinx sin y
= (cos2x + sin2x) + (cos2y + sin2y)  - 2[cos x cos y + sin x sin y]
= 1 + 1 - 2[cos (x - y)]   [cos(A - B) = cos A cos B + sin A sin B]
= 2[1 - cos (x - y)]
= 2[1 - {1 - 2sin2{(x -y)/2}]  [cos2A = 1 - 2 sin2 A]
= 4sin2{(x - y)/2} = R.H.S.

5.  Prove that sin x + sin 3x + sin 5x + sin 7x = 4 cosx cos2x sin4x
Solution
It is known that sin A + sin B = 2sin[(A + B)/2].cos [(A-B)/2]
∴ L.H.S = sin x + sin 3x + sin 5x + sin 7x
= (sin x + sin 5x) + (sin 3x + sin 7x)
= 2sin [(x + 5x)/2] .cos [(x - 5x)/2] + 2 sin [(3x + 7x)/2] cos [(3x - 7x)/2]
= 2 sin 3x cos(-2x) + 2 sin 5x cos (-2x)
= 2 sin 3x cos 2x + 2sin 5x cos 2x
= 2 cos 2x[sin 3x + sin 5x]
= 2 cos 2x[2sin{(3x + 5x)/2}. cos{(3x - 5x)/2}
= 2cos 2x[2sin4x. cos(-x)]
= 4 cos 2x sin 4x cos x = R.H.S.

6. Prove that : [(sin 7x + sin 5x) + (sin 9x + sin 3x)]/[(cos 7x + cos 5x)+(cos 9x + cos 3x)] = tan 6x
Solution
It is known that

= tan 6x
R.H.S

7. Prove that : sin 3x + sin 2x - sin x = 4sin x cos (x/2) cos (3x/2)
Solution
L.H.S. = sin 3x + sin 2x - sin x
= sin 3x + (sin 2x - sin x)

8. tan x = -4/3, x in quadrant II
Solution
Here, x is in quadrant II.
i.e., Ï€/2 < x < Ï€
⇒ Ï€/4 < x/2 < Ï€/2
Therefore, sin(x/2), cos (x/2) and tan(x/2) are all positive.
It is given that tan x = -4/3
sec2 x = 1 + tan2 x = 1 + (-4/3)2 = 1 + 16/9 = 25/9
∴ cos2 x = 9/25
⇒ cos x = ± (3/5)
As x is in quadrant II, cos x is negative.
∴ cos x = -3/5
Now, cos x = 2 cos2(x/2) - 1
⇒ -3/5 = 2cos2(x/2) - 1
⇒ 2cos2(x/2) = 1 - 3/5
⇒ 2cos2(x/2) = 2/5
⇒ cos2(x/2) = 1/5
⇒ cos(x/2) = 1/√5 [∵ cos(x/2) is positive]
∴ cos(x/2) = √5/5
sin2(x/2) + cos2(x/2) = 1
⇒ sin2 (x/2) + (1/√5)2  = 1
⇒ sin2 (x/2) = 1 - 1/5 = 4/5
⇒ sin (x/2) = 2/√5  [∵sin (x/2) is positive]
∴ sin (x/2) = 2√5/5

Thus, the respective values of sin(x/2), cos(x/2) and tan(x/2) are 2√5/5, √5/5, and 2.

9. Find sin(x/2), cos (x/2) and tan(x/2) for cos x = -1/3, x in quadrant III
Solution
Here, x is in quadrant III.
i.e., Ï€ < x < 3Ï€/2
⇒ Ï€/2 < x/2 < 3Ï€/4
Therefore, cos (x/2) and tan(x/2) are negative, whereas sin(x/2) is positive.
It is given that cos x = -1/3 .
cos x = 1 - 2sin2(x/2)
⇒ sin2(x/2) = (1 - cos x)/2

Thus, the respective values of sin(x/2), cos(x/2) and tan(x/2) are √6/3, -√3/2 and -√2.

10. Find, sin(x/2) cos(x/2) and tan(x/2) for sin x = 1/4 , x in quadrant II
Solution
Here, x is in quadrant II.
i.e., Ï€/2 < x < Ï€
⇒ Ï€/4 < x/2 < Ï€/2
Therefore, sin (x/2), cos (x/2), and tan(x/2) are all positive.
It is given that sin x = 1/4.
cos2x = 1 - sin2x  = 1 - (1/4)2 = 1 - 1/16 = 15/16
⇒ cos x = -√15/4  [cos x is negative in quadrant II]

Thus, the respective values of sin(x/2), cos(x/2) and tan(x/2) are and  4+√15.