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}).