I'm trying to find the closed form of the above formula. This link shows the solution of tan version. Sum of tangent functions where arguments are in specific arithmetic series Though I'm trying to find the way to apply the argument used in the link to this case, it becomes quite complicated. Could you show another method to get the formula or modify my method? Do you think your formula is similar to the closed formula of sum of cos's?

My method:

Using De Moivre's formula,

$$\sin rx=\binom r1\cos^{r-1}x\sin x-\binom r1\cos^{r-3}x\sin^3x+\binom r5\cos^{r-5}x\sin^5x-\cdots$$ $$=\cos^nx\left(\binom r1 \tan x-\binom r1\tan^3x+\binom r5\tan^5x-\cdots\right)$$ and

If $r$ is even, $=2m$(say) $$\sin2m\theta=\sin2mx=\cos^n{x}\left(\binom {2m}1\tan x-\binom {2m}3\tan^3x+\cdots+\binom {2m}{2m-1}(-1)^{m-1}\tan^{2m-1}x\right)$$

If $n$ is even, I can convert $$\cos^nx=\frac{1}{1+\tan^{n/2}x}$$

and if I investigate each case of the parity of $m$, I can get the formula of the case in which $r$ is even. But otherwise, I can't. Also, I have to investigate the case in which $n$ is odd.

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    $\begingroup$ If you use Euler's formula, you can write $\sum \sin (k\alpha + \theta) = \operatorname{Im} \sum e^{i(k\alpha + \theta)} = \operatorname{Im} \left(e^{i\theta}\sum e^{ik\alpha}\right)$. Then the sum is a geometric sum, which is relatively easy to handle. $\endgroup$ – Daniel Fischer Nov 4 '13 at 21:31
  • $\begingroup$ Riemann sum requires $\Delta{x}$ to be converted to integration. In this case, $\Delta{x}=\frac{1}{n}$. So, $\sum_{k=0}^{n-1}\frac{1}{n}\sin(\frac{k}{n}+\theta)=\int_0^1 \sin(x\pi+\theta)dx$ But our summation doesn't have the $\Delta x$, so your integration becomes infinite. $\endgroup$ – Math.StackExchange Nov 5 '13 at 1:14
  • $\begingroup$ Yes, $\frac1n$ was missing. (I've only just now noticed it and came back here to delete my silly comment). Thank you anyway for taking the time to respond ! :-) $\endgroup$ – Lucian Nov 5 '13 at 2:02

For sums of sines (and cosines) where the arguments are in an arithmetic progression, we have a much simpler way using Euler's formula:

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

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  • $\begingroup$ OK. So, this method is not applicable to the sum of tan's or cot's but to the sum of sin's, cos's, csc's, or sec's? $\endgroup$ – Math.StackExchange Nov 4 '13 at 22:34
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    $\begingroup$ Only sines and cosines. For $\sec$ or $\csc$, you must do something similar to the $\tan$ method. For $\sec$, for example, the terms are $\dfrac{2}{e^{ikx}+e^{-ikx}}$, and you can't split that nicely into a geometric sum as you can do for $\sin$ and $\cos$. $\endgroup$ – Daniel Fischer Nov 4 '13 at 22:40
  • $\begingroup$ I see. Thanks for your extra help. $\endgroup$ – Math.StackExchange Nov 5 '13 at 1:03


$$\sum_{k=0}^{n-1} \sin{\left ( k \frac{\pi}{n} + \theta\right)} = \Im{\left [e^{i \theta} \sum_{k=0}^{n-1} e^{i k \pi/n} \right ]}$$

Now, use a geometric series and a half-angle formula to show that

$$\sum_{k=0}^{n-1} e^{i k \pi/n} = \frac{1-e^{-i \pi/n}}{2 \sin^2{\left ( \frac{\pi}{2 n}\right)}}$$

Then you will come across several trig identities in showing that

$$\sum_{k=0}^{n-1} \sin{\left ( k \frac{\pi}{n} + \theta\right)} = \frac{1}{2 \sin^2{\left ( \frac{\pi}{2 n}\right)}} \Im{\left [e^{i \theta} \left ( 1-e^{-i \pi/n}\right) \right ]} = \frac{\cos{\left ( \theta-\frac{\pi}{2 n}\right)}}{\sin{\frac{\pi}{2 n}}}$$

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