Asked by Mahlatse

Prove that: (sinx-2sin2x+sin3x)/(sinx+2sin2x+sin3x)= -tan^2 x/2.
Answered by Steve
sinx + sin3x = 4sinx cos^2x = 2sin2x cosx

So, you have

(2sin2x cosx - 2sin2x)
---------------------------
(2sin2x cosx + 2sin2x)

2sin2x(cosx-1)
---------------------
2sin2x(cosx+1)

= (cosx-1)/(cosx+1)

Now take a look at the half-angle formula for tan(x/2)
Answered by Mahlatse
I don't understand it....
Answered by Steve
tan^2(x/2) = (1-cosx)/(1+cosx)
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