There aren't any bounds, $n$ is an even integer. I have no idea where to begin.


Hint: $\sin(z) = \frac{e^{iz}-e^{-iz}}{2i}$, hence:

$$ \frac{\sin(2kx)}{\sin x} = \frac{e^{2kix}-e^{-2kix}}{e^{ix}-e^{-ix}}=\sum_{j=0}^{2k-1}e^{jix}\cdot e^{-(2k-1-j)ix} $$ and the last sum is easy to integrate. As an alternative, you may prove that: $$ \frac{\sin(2kx)}{\sin(x)} = 2\cos(x)+2\cos(3x)+\ldots + 2\cos((2k-1)x) $$ since the RHS, multiplied by $\sin(x)$, becomes a telescopic sum.

  • $\begingroup$ How would I integrate the summation? $\endgroup$ – Dandy Don Nov 2 '15 at 12:41
  • $\begingroup$ @DandyDon: how do you usually integrate $\cos(n x)$? $\endgroup$ – Jack D'Aurizio Nov 2 '15 at 12:48
  • $\begingroup$ Okay got that, how would you start the proof for sin(2kx)/sin(x) = 2cos(x) +... $\endgroup$ – Dandy Don Nov 2 '15 at 14:00
  • $\begingroup$ @DandyDon: it is written above. $2\cos(3x)\sin(x) = \sin(4x)-\sin(2x)$, for instance. $\endgroup$ – Jack D'Aurizio Nov 2 '15 at 17:21

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