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I was given a task to calculate this sum: $$S=1+\frac{\sin x}{\sin x}+\frac{\sin 2x}{\sin^2 x}+\cdots +\frac{\sin nx}{\sin^n x}$$ but I'm not really sure how to start solving it. Like always, I would be grateful if someone could provide a subtle hint on how to begin. Thanks ;)

Michael Hardy
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Transcendental
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    Can you use de moive theorem and taking the imaginary part? ? – Chinny84 Dec 02 '14 at 16:51
  • I would get a common denominator and work with trig identities on the numerator. I'm not sure if that will work, but it seems a reasonable place to start. – KSmarts Dec 02 '14 at 16:54

2 Answers2

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Define $S^{'}=S-1$

Let $C^{'}=\dfrac{\cos x}{\sin x} + \dfrac{\cos2x}{\sin^2x} + ....... + \dfrac{\cos nx}{\sin^nx}$

Now,

$S^{'}=\Im{(C^{'}+iS^{'})}=\dfrac{e^{ix}}{\sin x} + \dfrac{e^{2ix}}{\sin^2x} + ...... + \dfrac{e^{nix}}{\sin^nx}$

Now sum it as a G.P. and get the imaginary part of the sum and lastly, add $1$.

MathGod
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Hint:

$\dfrac{\sin ((n+1)x)}{\sin^{n+1} x}=\dfrac{\sin (nx + x)}{\sin^n x \times \sin x}= \dfrac{\sin nx \times \cos x + \sin x \times \cos nx}{\sin^n x \times \sin x}$

Martigan
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