NCERT Solutions For Class 11 Math Chapter – 3 Miscellaneous

Miscellaneous Exercise page: 81

Prove that:

1.

Solution:

We get

= 0

= RHS

2. (sin 3x + sin x) sin x + (cos 3x – cos x) cos x = 0

Solution:

Consider

LHS = (sin 3x + sin x) sin x + (cos 3x – cos x) cos x

By further calculation

= sin 3x sin x + sin² x + cos 3x cos x – cos² x

Taking out the common terms

= cos 3x cos x + sin 3x sin x – (cos² x – sin² x)

Using the formula

cos (A – B) = cos A cos B + sin A sin B

= cos (3x – x) – cos 2x

So we get

= cos 2x – cos 2x

= 0

= RHS

3.

Solution:

Consider

LHS = (cos x + cos y) ² + (sin x – sin y) ²

By expanding using formula we get

= cos² x + cos² y + 2 cos x cos y + sin² x + sin² y – 2 sin x sin y

Grouping the terms

= (cos² x + sin² x) + (cos² y + sin² y) + 2 (cos x cos y – sin x sin y)

Using the formula cos (A + B) = (cos A cos B – sin A sin B)

= 1 + 1 + 2 cos (x + y)

By further calculation

= 2 + 2 cos (x + y)

Taking 2 as common

= 2 [1 + cos (x + y)]

From the formula cos 2A = 2 cos² A – 1

4.

Solution:

LHS = (cos x – cos y) ² + (sin x – sin y) ²

By expanding using formula

= cos² x + cos² y – 2 cos x cos y + sin² x + sin² y – 2 sin x sin y

Grouping the terms

= (cos² x + sin² x) + (cos² y + sin² y) – 2 (cos x cos y + sin x sin y)

Using the formula cos (A – B) = cos A cos B + sin A sin B

= 1 + 1 – 2 [cos (x – y)]

By further calculation

= 2 [1 – cos (x – y)]

From formula cos 2A = 1 – 2 sin² A

5. sin x + sin 3x + sin 5x + sin 7x = 4 cos x cos 2x sin 4x

Solution:

6.

Solution:

7.

Solution:

8. Find sin x/2, cos x/2 and tan x/2 in each of the following:

Solution:

cos x = -3/5

From the formula

9. cos x = -1/3, x in quadrant III

Solution:

10. sin x = 1/4, x in quadrant II

Solution: