1

$$\sin(\theta)+ \sin (2\theta)+\ldots+\sin(n\theta).$$

Find the identity? Set $z=e^{i\theta}$.

Can someone point me in the right direction as to where to start/go from.

So far I have

$$\frac{1-e^{i\theta(n+1)}}{1-e^{i\theta}}.$$

And I am stuck as to where to go from here.

okocj
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2 Answers2

0

$$\sum_{k=1}^ne^{ki\theta}=e^{i\theta}\frac{e^{ni\theta}-1}{e^{i\theta}-1}=\left(\cos\theta+i\sin\theta\right)\frac{\cos n\theta-1+i\sin n\theta}{\cos\theta-1+i\sin\theta}$$

Find now real and imaginary parts and complete.

Timbuc
  • 34,191
  • thank you for the start but just wondering why the 1 changed to -1 and the e^(itheta) outside – okocj Feb 18 '15 at 18:58
  • shouldnt it be e^itheta(e^(-itheta) - e^(itheta*n)/(1-e^(itheta))? – okocj Feb 18 '15 at 19:01
  • @O No, it is as I wrote it. – Timbuc Feb 18 '15 at 19:09
  • but the partial sum of 1 + z + z^2 + ... = (1-z^(n+1))/(1-z). I'd really appreciate if you could point where I am wrong. In the above solution you gave though, you multiplied by a minus. Can't I still use e^itheta(e^(-itheta) - e^(itheta*n)/(1-e^(itheta))?! – okocj Feb 18 '15 at 19:27
  • @okocj where do you take that minus in the exponent from?! And also pay attention to the fact that $$\frac{a-1}{b-1}=\frac{1-a}{1-b}$$ – Timbuc Feb 18 '15 at 20:25
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Here is a start $$ \begin{align} \sum_{k=1}^n\sin(k\theta) &=\frac1{2i}\sum_{k=1}^n\left(e^{ik\theta}-e^{-ik\theta}\right)\tag{1}\\ &=\frac1{2i}\left(\frac{e^{i(n+1)\theta}-e^{i\theta}}{e^{i\theta}-1} -\frac{e^{-i(n+1)\theta}-e^{-i\theta}}{e^{-i\theta}-1}\right)\tag{2}\\ &=\frac1{2i}\left(\frac{e^{i(n+1/2)\theta}-e^{i\theta/2}}{e^{i\theta/2}-e^{-i\theta/2}} -\frac{e^{-i(n+1/2)\theta}-e^{-i\theta/2}}{e^{-i\theta/2}-e^{i\theta/2}}\right)\tag{3} \end{align} $$ Explanation:
$(1)$: write $\sin(x)=\frac{e^{ix}-e^{-ix}}2$
$(2)$: sum of a geometric series
$(3)$: adjust to make the denominators similar

Here is the rest $$ \begin{align} \sum_{k=1}^n\sin(k\theta) &=\frac1{2i}\left(\frac{e^{i(n+1/2)\theta}-e^{i\theta/2}}{e^{i\theta/2}-e^{-i\theta/2}} -\frac{e^{-i(n+1/2)\theta}-e^{-i\theta/2}}{e^{-i\theta/2}-e^{i\theta/2}}\right)\\ &=\frac1{2i}\left(\frac{e^{i(n+1/2)\theta}+e^{-i(n+1/2)\theta}}{e^{i\theta/2}-e^{-i\theta/2}} -\frac{e^{i\theta/2}+e^{-i\theta/2}}{e^{i\theta/2}-e^{-i\theta/2}}\right)\\ &=\frac1{2i}\left(\frac{2\cos((n+1/2)\theta)}{2i\sin(\theta/2)} -\frac{2\cos(\theta/2)}{2i\sin(\theta/2)}\right)\\ &=\frac{\cos(\theta/2)-\cos((2n+1)\theta/2)}{2\sin(\theta/2)}\\ &=\frac{\sin((n+1)\theta/2)\sin(n\theta/2)}{\sin(\theta/2)} \end{align} $$

robjohn
  • 345,667
  • thanks alot a lot got it using this approach but the numerator was very messy...and my answer was (sin((n+1)/2)theta)(cosntheta/2))/sin(theta/2) – okocj Feb 18 '15 at 22:42
  • The answer I got from $(3)$ was $$\frac{\cos(\theta/2)-\cos((2n+1)\theta/2)}{2\sin(\theta/2)} =\frac{\sin((n+1)\theta/2)\sin(n\theta/2)}{\sin(\theta/2)}$$ – robjohn Feb 18 '15 at 22:55
  • @okocj: since you seem to have done the work with a small error, I will finish the answer. – robjohn Feb 19 '15 at 00:20