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Attempt at part b:

$\sin(\theta)+\sin(2\theta)+... = \frac {(e^{i\theta}-e^{-i\theta})+(e^{2i\theta}-e^{-2i\theta})+...+(e^{ni\theta}-e^{-ni\theta})} {2i} \\ = \frac {(1+e^{i\theta}+(e^{i\theta})^2+...+(e^{i\theta})^n) - (1+e^{-i\theta}+(e^{-i\theta})^2+...+(e^{-i\theta})^n)} {2i} \\ = \frac {\frac{(e^{i\theta})^{n+1}-1} {e^{i\theta}-1} - \frac {(e^{-i\theta})^{n+1}-1} {e^{-i\theta}-1}} {2i} \\ = \frac {(e^{i\theta n} - e^{i\theta(n+1)} - e^{-i\theta} + 1) - (e^{-i\theta n} - e^{-i\theta(n+1)} - e^{i\theta} + 1)} {2i(e^{i\theta} - 1)(e^{-i\theta}-1)} \\ = \frac {\cos(n\theta)+i\sin(n\theta) - (\cos((n+1)\theta)+i\sin((n+1)\theta)) - (\cos(\theta) - i\sin(\theta)) - (\cos(n\theta)-i\sin(n\theta) - (\cos((n+1)\theta)-i\sin((n+1)\theta)) - (\cos(\theta) + i\sin(\theta)))} {2i(1-(\cos(\theta) + i\sin(\theta)) - (\cos(\theta) - i\sin(\theta)) + 1)} \\ = \frac {\sin(n\theta) - \sin((n+1)\theta) + \sin(\theta)} {2(1 - \cos(\theta))} $ \

What to do from here?

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    $\begingroup$ Not a direct answer to yout question, but it could be more clean to work in the complex plane with $z=e^{i\theta}$ and then compute the real and imaginary part of the expression obtained. You can use this as a guide. $\endgroup$
    – rebo79
    Commented May 13, 2021 at 6:40
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    $\begingroup$ The question itself has been asked before here. As for completing the proof from where you are, use sum-to-product formulas in the numerator, and a half-angle formula in the denominator. $\endgroup$
    – Blue
    Commented May 13, 2021 at 6:42
  • $\begingroup$ I got $4\sin^2(\theta/2)$ in the denominator. But which terms do I combine in the numerator? I tried all combinations of 2 and couldn't find a continuation. $\endgroup$
    – Daniel
    Commented May 13, 2021 at 13:11

1 Answer 1

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$$\sin\theta + \sin2\theta + ... + \sin n\theta = \Im(1 + e^{i\theta} + e^{i2\theta} + \ldots + e^{in\theta}) $$ where $\Im (z)$ denotes the imaginary part of $z\in\mathbb C$.

$1 + e^{i\theta} + e^{i2\theta} + \ldots + e^{in\theta}$ is a geometric progresion with common ratio $e^{i\theta}$ and first term unity. I'm sure you can compute that $$1 + e^{i\theta} + e^{i2\theta} + \ldots + e^{in\theta} = \frac{1 - e^{i(n+1)\theta}}{1-e^{i\theta}}$$ Then, $$\sin\theta + \sin2\theta + ... + \sin n\theta = \Im\left(\frac{1 - e^{i(n+1)\theta}}{1-e^{i\theta}} \right)$$ Now, I'd ask you to use $e^{i\theta} = \cos\theta + i\sin\theta$ and half-angle trigonometric identities to simplify the denominator. You should be able to figure out $$\Im\left(\frac{1 - e^{i(n+1)\theta}}{1-e^{i\theta}} \right) = \frac {\sin(n\theta/2)\sin((n+1)\theta/2)} {\sin(\theta/2)}$$ If you face any trouble, let me know.

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  • $\begingroup$ I also tried doing it that way but got to the exact same point $\frac {\sin(n\theta) - \sin((n+1)\theta) + \sin(\theta)} {2(1 - \cos(\theta))}$ $\endgroup$
    – Daniel
    Commented May 14, 2021 at 6:34
  • $\begingroup$ After this point, you should use $\sin A + \sin B = 2\sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$ on $\sin n\theta + \sin ((-n-1)\theta)$, and the usual half angle identities for $\sin\theta$ and $1- \cos\theta$. $\endgroup$ Commented May 14, 2021 at 7:08

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