# GSEB Solutions Class 11 Maths Chapter 3 Trigonometric Functions Miscellaneous Exercise

Gujarat Board GSEB Textbook Solutions Class 11 Maths Chapter 3 Trigonometric Functions Miscellaneous Exercise Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 11 Maths Chapter 3 Trigonometric Functions Miscellaneous Exercise

1. 2cos $$\frac{Ï€}{13}$$cos $$\frac{9Ï€}{13}$$ + cos $$\frac{3Ï€}{13}$$ + cos $$\frac{5Ï€}{13}$$ = 0
2. (sin 3x + sin x)sin x + (cos 3x – cos x)cos x = 0
3. (cos x + cos y)2 + (sin x – sin y)2 = 4 cos2 $$\frac{x+y}{2}$$
4. (cos x – cos y)2 + (sin x – sin y)2 = 4 sin2 $$\frac{x-y}{2}$$
Solutions to questions 1 – 4:
1. L.H.S. = 2 cos $$\frac{Ï€}{13}$$cos $$\frac{9Ï€}{13}$$ + cos $$\frac{3Ï€}{13}$$ + cos $$\frac{5Ï€}{13}$$
= cos $$\frac{10Ï€}{13}$$ + cos $$\frac{8Ï€}{13}$$ + cos $$\frac{3Ï€}{13}$$ + cos $$\frac{5Ï€}{13}$$
[âˆµ 2cos A cos B = cos (A + B) + cos (A – B)]

= 0 = R.H.S.

2. L.H.S. = (sin 3x + sinx)sinx + (cos3x – cosx)cosx
= sin3xsinx + sin2x + cos3x cosx – cos2x
= (cos 3x cos x + sin 3x sin x) – (cos2x – sin2 x)
= cos (3x – x) – cos 2x + cos2x = cos 2x – cos 2x = 0
= R.H.S.

3. L.H.S. = (cos x + cos y)2 + (sin x – sin y)2

4. L.H.S. = (cos x – cos y)2 + (sin x – sin y)2

Prove that:
5. sin x + sin 3x + sin 5x + sin 7x = 4cos x cos 2x sin 4x
6. $$\frac{(sin 7x + sin 5x) + (sin 9x + sin 3x)}{(cos 7x + cos 5x) + (cos 9x + cos 3x)}$$ = tan 6x
7. sin 3x + sin 2x – sin x = 4sin x cos $$\frac{x}{2}$$ cos $$\frac{3x}{2}$$
solutions:
5. L.H.S. = sin x + sin 3x + sin 5x + sin 7x
= (sin 7x + sin x) + (sin 5x + sin 3x)
= 2sin $$\frac{7x + x}{2}$$cos $$\frac{7x – x}{2}$$ + 2sin $$\frac{5x + 3x}{2}$$cos $$\frac{5x – 3x}{2}$$
= 2sin 4x cos 3x + 2 + 2 sin 4x cos x
= 2sin 4x[cos 3x + cos x]
= 2sin 4x. 2cos $$\frac{3x+x}{2}$$cos $$\frac{3x-x}{2}$$
= 4 sin 4x. cos 2x. cos x
= 4cos x cos 2x. sin 4x = R.H.S.

6. L.H.S. =

7. L.H.S. = sin 3x + (sin 2x – sin x)

Find sin $$\frac{x}{2}$$, cos $$\frac{x}{2}$$ and tan $$\frac{x}{2}$$ in each of the following problems:
8. tan x = – $$\frac{4}{3}$$, x in quadrant II.
9. cos x = – $$\frac{1}{3}$$, x in quadrant III.
10. sin x = $$\frac{1}{4}$$, x in quadrant II.
Solutions to questions (8 – 10):
8. since x lies in the second quadrant, therefore cos x is negative.

Now x, lies in 2nd quadrant
â‡’ $$\frac{Ï€}{2}$$ < x < Ï€ â‡’ $$\frac{Ï€}{4}$$ < $$\frac{x}{2}$$ < $$\frac{Ï€}{2}$$
â‡’ $$\frac{x}{2}$$ lies in first quadrant.
â‡’ sin $$\frac{x}{2}$$, cos $$\frac{x}{2}$$ and tan $$\frac{x}{2}$$ are positive.

Since x lies in quadrant III, therefore
180Â° < x < 270Â°
â‡’ 90Â° < $$\frac{x}{2}$$ < 135Â°
or $$\frac{x}{2}$$ lies in quadrant II.

Since x lies in quadrant II, therefore
cos x < 0.

But x lies in quadrant II.

But cos $$\frac{x}{2}$$ > 0 for 45Â° â‰¤ $$\frac{x}{2}$$ â‰¤ 90Â°

= 4 + $$\sqrt{15}$$.