I have to find this sum: $$S= \sin(2x)+\sin(4x)+\sin(6x)+\cdots+\sin(22x).$$ I tried to multiply the $S$ by $i$ then add a so called "$T$" where $$T = \cos 2x + \cos 4x +\cdots+\cos 22x.$$ From here I obtained: $$T+iS=(\cos x+i\sin x)^2+(\cos 2x+i\sin2x)^2+\cdots+(\cos11x+i\sin11x)^2.$$ What should I do next?
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The trick is to multiply $S$ by $2\sin(x)$. – SMM May 17 '18 at 12:42
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Can you be more specific? – Robert Maracine May 17 '18 at 12:43
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Another approach is to note $\sin u=\Im e^{iu}$. – BAI May 17 '18 at 12:44
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@RobertMaracine Multiply $S$ by $2\sin(x)$, and then use the formula $2\sin(x)\sin(y)= \cos(y-x)-\cos(y+x)$. Almost all summands will vanish. – SMM May 17 '18 at 12:46
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@SMM Thanks a lot ! – Robert Maracine May 17 '18 at 12:54
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Why is each term squared in $T+iS$ – Abhishek Choudhary May 17 '18 at 13:22
1 Answers
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From your attempt $$T+iS=\left(\cos2x+i\sin2x\right)+\left(\cos4x+i\sin4x\right)...$$
From euler's formula $$e^{ix}=\cos{x}+i\sin{x}$$
$T+iS$ can be written as
$$e^{2ix}+e^{4ix}...$$
It is a G.P. with $a=e^{2ix}$ and $r=e^{2ix}$ and $n=11$,
$$\therefore T+iS=\frac{a\left(r^n-1\right)}{r-1}=\frac{e^{2ix}\left(e^{2nix}-1\right)}{e^{2ix}-1}$$
I think you should work it out yourself from here by expanding it and comparing real and imaginary parts
