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I have worked something out like the following

https://math.stackexchange.com/a/1172454/626039 - reproduced here: $$\begin{align} \sum_{k=1}^{n} \sin (k\theta)&=\Im \sum_{k=1}^{n} e^{ik\theta}\\\\ &=\Im\left( e^{i\theta}\frac{e^{in\theta}-1}{e^{i\theta}-1}\right)\\\\ &=\Im\left( e^{i\theta}\frac{e^{in\theta/2}\left(e^{in\theta/2}-e^{-in\theta/2}\right)}{e^{i\theta/2}\left(e^{i\theta/2}-e^{-i\theta/2}\right)}\right)\\\\ &=\Im\left( e^{i\theta}\frac{e^{in\theta/2}\left(2i\sin(n\theta/2)\right)}{e^{i\theta/2}\left(2i\sin(\theta/2)\right)}\right)\\\\ &=\Im\left( e^{i(n+1)\theta/2}\frac{\sin(n\theta/2)}{\sin(\theta/2)}\right)\\\\ &=\Im\left( \left(\cos ((n+1)\theta/2)+i\sin ((n+1)\theta/2)\right)\frac{\sin(n\theta/2)}{\sin(\theta/2)}\right)\\\\ &=\frac{\sin ((n+1)\theta/2)}{\sin(\theta/2)}\sin(n\theta/2). \end{align}$$

The only problem is that i want to do this for when the sequence starts at k = 0 NOT k = 1 . I can't seem to achieve the same result but it should do according to my question.

Paul Sinclair
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Adam Suttle
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  • CAS says: $\sum _{k=0}^n \sin (k \theta )=\csc \left(\frac{\theta }{2}\right) \sin \left(\frac{n \theta }{2}\right) \sin \left(\frac{1}{2} (1+n) \theta \right)$ – Mariusz Iwaniuk Apr 04 '19 at 17:01
  • FYI - if you right-click on a math equation here and select "Show Math As > TeX Commands", it will show you the TeX commands used to produce the mathematics, where you can copy them and paste them into your own post (inside the MathJax indicators $$ $$ to make it display). This is a much better option than taking snapshots and linking to them. – Paul Sinclair Apr 04 '19 at 23:43
  • But starting from 0 doesn’t change much, since $\sin$ vanishes at 0 – Fede Poncio Apr 04 '19 at 23:51

1 Answers1

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$$\begin{align} \sum_{k=0}^{n} \sin (k\theta)&=\Im \sum_{k=0}^{n} e^{ik\theta}\\\\ &=\Im\left(\frac{e^{i(n+1)\theta}-1}{e^{i\theta}-1}\right)\\\\ &=\Im\left(\frac{e^{i(n+1)\theta/2}\left(e^{i(n+1)\theta/2}-e^{-i(n+1)\theta/2}\right)}{e^{i\theta/2}\left(e^{i\theta/2}-e^{-i\theta/2}\right)}\right)\\\\ &=\Im\left(\frac{e^{i(n+1)\theta/2}\left(2i\sin((n+1)\theta/2)\right)}{e^{i\theta/2}\left(2i\sin(\theta/2)\right)}\right)\\\\ &=\Im\left(e^{in\theta/2}\frac{\sin((n+1)\theta/2)}{\sin(\theta/2)}\right)\\\\ &=\Im\left( \left(\cos (n\theta/2)+i\sin (n\theta/2)\right)\frac{\sin((n+1)\theta/2)}{\sin(\theta/2)}\right)\\\\ &=\frac{\sin ((n+1)\theta/2)}{\sin(\theta/2)}\sin(n\theta/2). \end{align}$$

As one would expect, as the term added for $k = 0$ is just $0$.

Paul Sinclair
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