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 $\frac{π}{13}$cos $\frac{9π}{13}$ + cos $\frac{3π}{13}$ + cos $\frac{5π}{13}$ = 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 $\frac{π}{13}$cos $\frac{9π}{13}$ + cos $\frac{3π}{13}$ + cos $\frac{5π}{13}$
= cos $\frac{10π}{13}$ + cos $\frac{8π}{13}$ + cos $\frac{3π}{13}$ + cos $\frac{5π}{13}$
[∵ 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
⇒ $\frac{π}{2}$ < x < π ⇒ $\frac{π}{4}$ < $\frac{x}{2}$ < $\frac{π}{2}$
⇒ $\frac{x}{2}$ lies in first quadrant.
⇒ sin $\frac{x}{2}$, cos $\frac{x}{2}$ and tan $\frac{x}{2}$ are positive.

Since x lies in quadrant III, therefore
180° < x < 270°
⇒ 90° < $\frac{x}{2}$ < 135°
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 45° ≤ $\frac{x}{2}$ ≤ 90°

= 4 + $\sqrt{15}$.