Very hard trig solution that is way too long

[tdude]

Homework Statement

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

Homework 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)

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?

 

[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.

 

[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

 

[tdude]

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

are you sure that's correct?

that's what's on wikipedia.

if it is however, it could help.

 

[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

 

[tdude]

aight, got it. all is well! Close thread please!