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I wonder if there is a way to calculate the

$$S_n=\sin x + \sin 2x + … + \sin nx$$

but using only derivatives ?

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    $\begingroup$ What do you mean by "using only derivatives"? $\endgroup$ Mar 2, 2015 at 20:50

2 Answers 2

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Using telescopic sums:

$$ \sin(mx)\sin(x/2) = \frac{1}{2}\left(\cos\left((m-1/2)x\right)-\cos\left((m+1/2)x\right)\right)$$ Hence: $$ S_n \sin\frac{x}{2} = \frac{1}{2}\left(\cos\frac{x}{2}-\cos\left(\left(n+\frac{1}{2}\right)x\right)\right)=\sin\frac{nx}{2}\cdot\sin\frac{(n+1)x}{2}.$$

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    $\begingroup$ Doesn't answer the question. $\endgroup$
    – abnry
    Mar 2, 2015 at 20:50
  • $\begingroup$ @nayrb: maybe, but this is easier than using derivatives. We just used... nothing. $\endgroup$ Mar 2, 2015 at 20:51
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    $\begingroup$ Still it is worth to be here. One can't be not amazed by this solution. Sorry for double negation. $\endgroup$
    – Mihail
    Mar 2, 2015 at 20:52
  • $\begingroup$ Okay, apparently it's what the OP wanted. $\endgroup$
    – abnry
    Mar 2, 2015 at 20:55
  • $\begingroup$ @JackD'Aurizio Astute! +1. $\endgroup$ Mar 2, 2015 at 20:56
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You may write $$ \begin{align} \sum_{k=1}^{n} \sin (k\theta)&=\Im \sum_{k=1}^{n} e^{ik\theta}\\\\ &=\Im\left( e^{i\theta}\frac{e^{in\theta}-1}{e^{i\theta}-1}\right)\\\\ &=\Im\left( e^{i\theta}\frac{e^{in\theta/2}\left(e^{in\theta/2}-e^{-in\theta/2}\right)}{e^{i\theta/2}\left(e^{i\theta/2}-e^{-i\theta/2}\right)}\right)\\\\ &=\Im\left( e^{i\theta}\frac{e^{in\theta/2}\left(2i\sin(n\theta/2)\right)}{e^{i\theta/2}\left(2i\sin(\theta/2)\right)}\right)\\\\ &=\Im\left( e^{i(n+1)\theta/2}\frac{\sin(n\theta/2)}{\sin(\theta/2)}\right)\\\\ &=\Im\left( \left(\cos ((n+1)\theta/2)+i\sin ((n+1)\theta/2)\right)\frac{\sin(n\theta/2)}{\sin(\theta/2)}\right)\\\\ &=\frac{\sin ((n+1)\theta/2)}{\sin(\theta/2)}\sin(n\theta/2). \end{align} $$

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    $\begingroup$ (+1) It was a mess to find the correct final form also to me :D $\endgroup$ Mar 2, 2015 at 20:51
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    $\begingroup$ @JackD'Aurizio That's what happens when I try to be fast ;). Thanks! $\endgroup$ Mar 2, 2015 at 20:54

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