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**differentiate** if u can do this, you are a genius

Prove by mathematical induction:

$\displaystyle \sum_{k=1}^{n}sink\theta = \frac{sin[\frac{1}{2}(n+1)\theta ]sin(\frac{1}{2}n\theta )}{sin(\frac{1}{2}\theta )} $

For n =1,

LHS = $\displaystyle sin\theta $

RHS = $\displaystyle \frac{sin(\frac{1}{2}(2\theta ))sin(\frac{1}{2})\theta }{sin(\frac{1}{2}\theta )} = sin\theta$

therefore, it is true for n =1

assume that it is true for n = a,

$\displaystyle \sum_{k=1}^{a}sina\theta = \frac{sin[\frac{1}{2}(a+1)\theta ]sin(\frac{1}{2}a\theta )}{sin(\frac{1}{2}\theta )} $

for n = a + 1

Prove: $\displaystyle \sum_{k=1}^{a}sina\theta = \frac{sin[\frac{1}{2}(a+2)\theta ]sin(\frac{1}{2}(a+1)\theta )}{sin(\frac{1}{2}\theta )} $

I don't know how to do the next step. I'm really bad with trigonometry.

HELP!

Thank you in advance