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How to prove that $\sum_{k=1}^{n} \sin{kx} = \frac{\cos{\frac{x}{2} - \cos{(n+\frac{1}{2})x}}}{2\sin{\frac{x}{2}}}$ for every positive integer $n$ where $x$ is a real number such that $\sin{\frac{x}{2}}\neq 0$?

My idea I am thinking of considering the sum $\sum_{k=1}^{n} 2\sin{kx} \sin{\frac{x}{2}}$ as an approach to this proof, but I am not sure how to proceed. Is my approach correct? Any tips or hints given? Thanks.

jjagmath
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unmyelinated neuron
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    Do you know about complex numbers and Euler's formula $e^{ikx}=\cos(kx)+i\sin(kx)$? This really is the most efficient way I think to approach the question (because for integer $k$, we have $e^{ikx}=(e^{ix})^k$, so summing it up is easy using the formula for a geometric sum). Alternatively, you can try induction and the various addition formulae. – peek-a-boo Mar 31 '22 at 01:34
  • btw the benefit of using the complex approach is that you can easily obtain the related identity $\sum_{k=1}^n\cos(kx)=\frac{\sin\left(n+\frac{1}{2}\right)x-\sin\left(\frac{x}{2}\right)}{2\sin\left(\frac{x}{2}\right)}$. – peek-a-boo Mar 31 '22 at 01:49
  • @metamorphy great reference thank you! I need a bit more time to digest all its contents haha – unmyelinated neuron Mar 31 '22 at 03:46
  • @peek-a-boo thank you for your suggestion! – unmyelinated neuron Mar 31 '22 at 03:47
  • @metamorphy Finally, I have proved it successfully. Thanks for sharing this link! – unmyelinated neuron Mar 31 '22 at 03:58

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