# Homework Help: Very hard trig solution that is way too long

1. Sep 21, 2009

### tdude

1. The problem statement, all variables and given/known data

Prove that
sin(5A)+sin(2A)-sin(A) = sin(2A)*(2*cos(3A)+1)

2. Relevant equations

sin(2A) = 2(sin(A)cos(A))
cos(2B) = cos2(A)-sin2(A)
cos(A+B) = cos(A)cos(B) - sin(A)sin(B)
sin(A+B) = sin(A)cos(B) + sin(B)cos(A)

3. The attempt at a solution

I can't find any other way than to just decompose the whole thing into sinA or cosA, and that is really long and usually full of mistakes. Is there something that i'm missing here? Something that makes this proof a lot easier?

2. Sep 21, 2009

### Dick

Well, yeah, somewhat easier. Expand the right side and cancel the sin(2A) on both sides. Now expand sin(5A)=sin(2A+3A). Then maybe use sin(3A)cos(2A)-cos(3A)sin(2A)=sin(A)? It goes a bit easier, yes? Hope I don't have a typo in there.

Last edited: Sep 21, 2009
3. Sep 21, 2009

### ehild

Do you know the identity

sin(A)*cos(B) = sin(A+B)+sin(A-B) ?

Use on the right side for 2*sin(2A)*cos(3A).

ehild

4. Sep 21, 2009

### tdude

are you sure that's correct?

that's what's on wikipedia.

if it is however, it could help.

5. Sep 22, 2009

### ehild

My mistake, I forgot the division by two. Use Wiki's formula, it really helps! You will be surprised, how easy the solution is. :)

ehild

6. Sep 22, 2009