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I was given an exercise:

Calculate 1+$\sum_{k=1}^{k=n}\frac{\sin(kx)}{\sin^{k}(x)}$

I recognize $$\sin(kx)=Im(cis(kx))=Im(cis^{k}(x))$$ and $$\sin^{k}(x)=(Im(cis(x)))^{k}$$ but I do not know how to proceed .

I would appreciate any help or hint on how to get started, I guess that is should be related to geometric series, but I didn't manage to get to any geometric series.

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What is $cis$? ${}{}$ –  no identity Oct 31 '12 at 14:15
    
@Norbert - $cis(\theta):=\cos(\theta)+i\sin(\theta)$ –  Belgi Oct 31 '12 at 14:16

1 Answer 1

up vote 3 down vote accepted

Hint:

  1. Prove that $$ \frac{\sin(kx)}{\sin^k(x)}=\frac{1}{2i}\left(\left(\frac{e^{i x}}{\sin x}\right)^k-\left(\frac{e^{-i x}}{\sin x}\right)^k\right) $$

  2. Recall geometric series formula.

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Thank you for your hint! –  Belgi Nov 1 '12 at 5:55

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