Question

The value of $si{n}^2$ B+ 4 cos (A + B) sin B + cos 2(A + B) is

• A. 0
• B. cos 3A
• C. cos 2A
• D. none of these

Answer 1

The correct option is C

cos 2A

We have,

2 $si{n}^2$ B + 4 cos (A + B) sin A sin B + cos 2(A + B)

= 1 - cos 2B + cos 2(A + B) + 4 cos (A + B) sin A sin B

= 1 + (cos2 (A + B) - cos 2B) + 4 cos (A + B) sin A sin B

$[\because cosC-cosD=-2sin\frac{C+D}{2}sin\frac{C-D}{2}]$

= 1 - 2 sin A [sin (A + 2B) - {sin (A + 2B) + sin (-A)}]

= 1 - 2 sin A [sin A]

= 1 - $2si{n}^2$ A

= cos 2A