# What is another form of cos(30◦-A)+ sin(60◦+A) ?

### 1 Answer | Add Yours

You need to remember that `sin alpha = cos(90^o - alpha)` , hence `sin (60^o + A) = cos(90^o - 60^o - A)`

`sin (60^o + A) = cos(30^o - A)`

Hence, substituting `cos(30^o - A)` for `sin (60^o + A)` in `cos(30^o - A) + sin (60^o + A)` yields:

`cos(30^o - A) + sin (60^o + A) = cos(30^o - A) + cos(30^o - A)`

`cos(30^o - A) + sin (60^o + A) = 2cos(30^o - A)`

You need to expand `cos(30^o - A)` such that:

`cos(30^o - A) = cos30^o*cos A + sin 30^o*sin A`

Substituting `1/2` for sin `30^o` and `sqrt3/2` for cos `30^o` yields:

`cos(30^o - A) = sqrt3*cos A/2 + sin A/2`

Hence, multiplying by 2 both sides yields:

`2cos(30^o - A) = sqrt3*cos A + sin A`

**Hence `cos(30^o - A) + sin (60^o + A) = sqrt3*cos A + sin A` .**