How can I prove that: $$\sin(x)+\sin(2x)+\cdots+\sin(nx)=\frac{\sin\left(\frac{x(n+1)}{2}\right)\sin\left(\frac{nx}{2}\right)}{\sin\left(\frac{x}{2}\right)}?$$
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You should post what you've tried, and what problems you had – Ivan Lerner Mar 11 '15 at 03:45
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I'm sure taking imaginary part of $\sum e^{inx}$, which is a geometric series, will go a long way... – Tim Clark Mar 11 '15 at 03:46
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I've tried with sin(mx)sin(nx)=1/2[cos((m-n)x)-cos((m+n)x)], but every time I try to get there I get confused. – ART Mar 11 '15 at 04:23
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let $$S = \sin x + \sin 2x + \cdots + \sin nx. $$ then if you multiply left hand side by $2\sin(x/2),$ you get $$\begin{align} 2\sin(x/2)S &= 2\sin(x/2) \sin x + 2\sin(x/2) \sin 2x + \cdots +2\sin(x/2) \sin(nx) \\ &= (\cos(x-x/2) - \cos(x+x/2)) + (\cos(2x-x/2) - \cos(2x+x/2) + \cdots (\cos(nx - x/2) - \cos(nx + x/2)\\ &= \cos(x/2) - \cos(nx + x/2)\\ &=2\sin(n+1)x/2\sin(nx/2)\end{align}$$
that is $$ S = \frac{\sin(n+1)x/2\sin(nx/2)}{\sin(x/2)}.$$
abel
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Thanks Abel. It made me look like a dumb, but, When does Math doesn't make one look dumb afterall? – ART Mar 11 '15 at 04:34
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Another simple way to solve the problem is to use complex numbers:
$$\sum_{k=1}^n\sin(kx)=\sum_{k=1}^n\Im(e^{ikx})=\Im\sum_{k=1}^ne^{ikx}=\Im(e^{ix}\frac{1-e^{inx}}{1-e^{ix}})\\=\Im(e^{ix+(inx/2)-(ix/2)}\frac{e^{-inx/2}-e^{inx/2}}{e^{-ix/2}-e^{ix/2}})\\=\Im(e^{i(n+1)x/2})\frac{\sin(nx/2)}{\sin(x/2)}=\frac{\sin((n+1)x/2)\sin(nx/2)}{\sin(x/2)}.$$
Here, $\Im(x+iy)=y$, for $x,y\in\mathbb R$.
Eclipse Sun
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Then the answer by abel is quite nice for you. Induction is also a good choice here. – Eclipse Sun Mar 11 '15 at 04:45