Question

_ _ _cos b =1/2[sin(a+b)+sin(a-b)]

A) cos b
B) sin b
C) sin a
D) cos a

Answer 1

sin a

Explanation:

let the unknown term be x.

`x.cos \: b = (1)/(2)(sin(a + b) + sin(a - b)) \\ or \: x = ( (1)/(2) 2sinacosb)/(cosb) \\ or \: x = sin \: a`

you have to remember that:

2sinAcosB=sin(A+B)+sin(A-B)