$\sum_{k=1}^n\sin kx=\frac{\cos\left(\frac{1}2x\right)-\cos\left(n+\frac{1}2\right)x}{2\sin(x/2)},x\neq0,\pm\pi,\pm 2\pi,\ldots$

Asked Jul 24 '18 at 16:48

Active Jul 24 '18 at 19:36

Viewed 29 times

Maybe I have a trivial question but , Could you tell me how did we get this fact please ?

$$\sum_{k=1}^n\sin kx=\frac{\cos\left(\frac{1}2x\right)-\cos\left(n+\frac{1}2\right)x}{2\sin(x/2)},x\neq0,\pm\pi,\pm 2\pi,\ldots$$

edited Jul 24 '18 at 19:36

Michael Hardy

asked Jul 24 '18 at 16:48

You are welcome. But maybe try to do a little bit of research within already answered or respectively asked question on this side before you ask by yourself :) – mrtaurho Jul 24 '18 at 16:58

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