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Sum of Trigonometric series

$\sin (x)-\sin(2x)+\sin(3x)-\sin(4x)+\cdots n$ terms

Try: I have take $2$ cases

$\bullet$ If $n$ is even natural number, Then

$S=\sin(x)-\sin(2x)+\sin(3x)+\cdots -\sin(nx)$

$\displaystyle 2S\cos\frac{x}{2}=\bigg(\sin \frac{3x}{2}+\sin \frac{x}{2}\bigg)-\bigg(\sin\frac{5x}{2}+\sin\frac{3x}{2}\bigg)+\bigg(\sin\frac{7x}{2}+\sin\frac{5x}{2}\bigg)+\cdots -\bigg(\sin (nx+\frac{x}{2})+\sin(nx-\frac{x}{2})\bigg)$

So $\displaystyle S=\frac{1}{2\cos \frac{x}{2}}\bigg(\sin \frac{x}{2}-\sin\bigg(nx\frac{x}{2}\bigg)\bigg)$

same way for $\bullet $ for $n$ is odd natural number

Could some help me? How can I solve it some less complex way, Thanks

Community
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DXT
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  • What is your question? –  Feb 25 '19 at 10:34
  • This approach https://math.stackexchange.com/questions/17966/how-can-we-sum-up-sin-and-cos-series-when-the-angles-are-in-arithmetic-pro/17967#comment39577_17967 should work in your case as well. – Martin R Feb 25 '19 at 10:41
  • Calling $z = e^{ix}$ we have

    $$ \sum_{k=1}^n (-1)^{k+1}z^k = -\frac{z^{n+1}+1}{z+1} $$

     – Cesareo Feb 25 '19 at 10:50

3 Answers3

2

You want $$ S = \sum_{k=1}^n (-1)^{k+1} \sin(kx) =\\ \Im (\sum_{k=1}^n (-1)^{k+1} e^{ikx}) $$ Using the finite geometric series (which is applicable for any $x$) we get $$ S = - \Im \frac{e^{ix(n+1)}+1}{e^{ix}+1} =\\ - \Im \frac{e^{ix(n+1/2)}+e^{-ix/2}}{e^{ix/2}+e^{-ix/2}} =\\ - \frac12 \Im \frac{e^{ix(n+1/2)}+e^{-ix/2}}{\cos(x/2)} =\\ \frac{1}{2\cos(x/2)}\left[-\sin(x(n+1/2)) + \sin(x/2) \right] $$

Andreas
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It's $$-2\sin\frac{x}{2}\left(\cos\frac{3x}{2}+\cos\frac{7x}{2}+...+\cos\frac{2n-1}{2}\right)=$$ $$=\frac{-2\sin\frac{x}{2}\left(2\sin{x}\cos\frac{3x}{2}+2\sin{x}\cos\frac{7x}{2}+...+2\sin{x}\cos\frac{2n-1}{2}\right)}{2\sin{x}}=$$ $$=-\frac{\sin\frac{5x}{2}-\sin\frac{x}{2}+\sin\frac{9x}{2}-\sin\frac{5x}{2}+...+\sin\frac{(2n+1)x}{2}-\sin\frac{(2n-3)x}{2}}{2\cos\frac{x}{2}}=$$ $$=-\frac{\sin\frac{(2n+1)x}{2}-\sin\frac{x}{2}}{2\cos\frac{x}{2}}=-\frac{\sin\frac{nx}{2}\cos\frac{(n+1)x}{2}}{\cos\frac{x}{2}}.$$

Michael Rozenberg
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If $\sin y=\sin x$

$y=2n\pi+x$ or $y=(2n+1)\pi-x$

Now for the first case $F=\sin my=\sin m(2n\pi+x)=\sin mx$ for any integer $m$

For the second case, $F=\sin my=\sin m(2n\pi+\pi-x)=\sin(m\pi-mx)$ for any integer $m$

$F=\sin mx$ for odd $m$

$F=-\sin mx$ for even $m$

So, we have $y=(2n+1)\pi-x$

WLOG $n=0, y=\pi-x$

We need $\sum_{r=1}^n\sin(ry)$

Use How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression?

lab bhattacharjee
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