<s>[INST] Sum of series $\sin x + \sin 2x + \sin 3x + \cdots $ Please help me compute the sum of the series:
 $$\sin(x)+\sin(2x)+\sin(3x)+\cdots$$
 [/INST] $$2\sin\frac{x}{2}\sin rx=\cos\frac{(2r-1)}{2}x-\cos\frac{(2r+1)}{2}x$$
Putting $r=1,2,\ldots,n-1,n$ we get, 
$$2\sin\frac{x}{2}\sin x=\cos\frac{1}{2}x-\cos\frac{3}{2}x$$
$$2\sin\frac{x}{2}\sin 2x=\cos\frac{3}{2}x-\cos\frac{5}{2}x$$
$$\vdots$$
$$2\sin\frac{x}{2}\sin rx=\cos\frac{(2n-3)}{2}x-\cos\frac{(2n-1)}{2}x$$
$$2\sin\frac{x}{2}\sin nx=\cos\frac{(2n-1)}{2}x-\cos\frac{(2n+1)}{2}x$$
Adding we get, $2\sin\frac{x}{2}(\sin x+\sin 2x+...+\sin nx)=cos\frac{1}{2}x-\cos\frac{(2n+1)}{2}x=2\sin\frac{(n+1)x}{2}\sin\frac{nx}{2}$
So, $$\sin x+\sin 2x+\cdots+\sin nx=\frac{\sin\frac{(n+1)x}{2}\sin\frac{nx}{2}}{\sin\frac{x}{2}}$$
As $2\sin B\sin(A+2rB) = \cos(A+(2r-1)B) - \cos(A+(2r+1)B)$, we need to multiply with $2\sin B$ for $\sum_{r}\sin(A+2rB)$.
Here in this problem, $A=0, 2B=x$

Also the way Nerd-Herd has approached the problem,
$\sin rx$ = Imaginary part of $e^{irx}$
So, $\sum_{0 ≤ r ≤n}\sin rx=\sum_{0 ≤ r ≤n}$(Imaginary part of $e^{ix})$=Imaginary part of($\sum_{0 ≤ r ≤n}e^{ix}$)
Now,
$$\sum_{0 ≤ r ≤n}e^{irx}= \frac{e^{(n+1)ix}-1}{e^{ix}-1}= e^{\frac{(n+1)ix}{2}}\frac{(e^{\frac{(n+1)ix}{2}}-e^{-\frac{(n+1)ix}{2}})}{e^{\frac{ix}{2}}(e^{\frac{ix}{2}}-e^{-\frac{ix}{2}})}$$
We know, $\sin y=\frac{e^{iy}-e^{-iy}}{2i}$ So,$e^{iy}-e^{-iy}=2i\sin y$
So,$\sum_{0 ≤ r ≤n}e^{irx}=e^{\frac{inx}{2}}\frac{2i\sin\frac{(n+1)x}{2}}{2i\sin\frac{x}{2}}=(\cos \frac{nx}{2}+i\sin \frac{nx}{2})\frac{\sin\frac{(n+1)x}{2}}{\sin\frac{x}{2}}$
So, imaginary part of $\sum_{0 ≤ r ≤n}e^{irx}=\sin \frac{nx}{2}\frac{\sin\frac{(n+1)x}{2}}{\sin\frac{x}{2}}$
$\sum_{0 ≤ r ≤n}\sin{rx}=\sin \frac{nx}{2}\frac{\sin\frac{(n+1)x}{2}}{\sin\frac{x}{2}}$
$\sum_{1 ≤ r ≤n}\sin{rx}=\sin \frac{nx}{2}\frac{\sin\frac{(n+1)x}{2}}{\sin\frac{x}{2}}$  as $\sin(0x)=0$
So, the either approach leads to a compact from provided $\sin\frac{x}{2}≠0$
If $\sin\frac{x}{2}=0$ i.e., $\frac{x}{2}=m\pi$ where m is some integer,
$=>x=2m\pi=>\sin sx= 0$ for any integer s.
By observation if $\sin x=0$ i.e., $x=m\pi$ where m is some integer, $\sin sx= 0$ 
for any integer s.
Then the sum is clearly 0 if $x=m\pi$.
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