Trying to prove a trig identity

[Appleton]

Homework Statement

prove that (sinA +sin3A + sin5A)/(cosA + cos3A + cos5A) = tan3A

Homework Equations

sinP + sinQ = 2sin((P+Q)/2)cos((P-Q)/2)
cosP + cosQ = 2cos((P+Q)/2)cos((P-Q)/2)

The Attempt at a Solution

(sin3A + sinA) + sin5A = 2sin2AcosA + 2sin((5/2)A)cos((5/2)A)
(cos3A + cosA) + cos5A = 2cos2AcosA + 2cos((5/2)A + 45)cos((5/2)A - 45)
It started to feel like a bit of a cul de sac at this point so I tried pursuing variations on this theme by starting with (sin5A + sin3A) + sinA etc, but these seemed just as fruitless.

 

[voko]

Because the right hand side is a function of 3A, it could be useful to combine A and 5A on the left hand side.

 

[verty]

I haven't checked what Voko suggested but you could also try this: let p = 3A, q = 2A.

 

[Saitama]

I haven't checked what Voko suggested but you could also try this: let p = 3A, q = 2A.

voko's suggestion is nice. I was able to solve the problem without pen and paper using his hint. I would suggest Appleton to try that. :)

 

[Appleton]

Thanks for all your suggestions, voko's suggestion led me to the proof, however, I'm still unable to eliminate the pen and paper.