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I need help to prove that the following is true:

$$1+\cos2\theta+\cos4\theta+\cos6\theta+\cos8\theta=\frac{(\cos4\theta)(\sin5 \theta)}{\sin\theta}$$

I realize that I must evaluate the real part of this, but whatever I get I am not quite sure how to get to the required expression. I have multiplied the numerator and denominator of the result of the geometric sum by the conjugate of $ e^{2i\theta} $ and still have no luck. $$\sum_{i=0}^4 e^{2ni\theta}$$

(Apologies for poor formatting)

rtybase
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Benjamin
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    That sum is a GP. – Angina Seng Jun 06 '18 at 18:33
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    @LordSharktheUnknown I did try to communicate that I was aware of this. Perhaps I did not phrase it well enough, but I have already used the geometric series formula. – Benjamin Jun 06 '18 at 18:39
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    Hint: don't multiply by the conjugate, but use half angles, to make trig functions appear. There remains a factor that is easier to simplify. That is: $$\dfrac{e^{2i(n+1)\theta}-1}{e^{2i\theta}-1}=\dfrac{e^{i(n+1)\theta}}{e^{i\theta}}\dfrac{\sin(n+1)\theta}{\sin\theta}$$ – Jean-Claude Arbaut Jun 06 '18 at 18:40
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    Once you have the sum of the GP, getting the real part is straightforward. You could say, add the expression to its conjugate. @Benjamin – Angina Seng Jun 06 '18 at 18:44

2 Answers2

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It isn't true (try putting $\theta = \pi/2$). It should be $\sin5\theta$ on the RHS. To prove it, you could multiply the LHS by $2\sin\theta$ and use $2\sin\theta\cos2\theta = \sin3\theta - \sin\theta$, etc. Most of the terms cancel and you end up with $\sin\theta + \sin9\theta = 2\cos4\theta\sin5\theta$.

Michael Behrend
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According to the link posted as a possible duplicate and containing my answer in it, we have: $$1+\cos{\theta}+\cos{2\theta}+...+\cos{n\theta}=\frac{\cos{\frac{n\theta}{2}}\cdot\sin{\frac{(n+1)\theta}{2}}}{\sin{\frac{\theta}{2}}} \tag{1}$$ Now $$1 + \cos{2\theta} + \cos{4 \theta} + \cos{6 \theta} + \cos8{\theta}=\\ 1 + \cos{(1\cdot \color{red}{2\theta})} + \cos{(2\cdot \color{red}{2\theta})} + \cos{(3 \cdot \color{red}{2\theta})} + \cos{(4\cdot \color{red}{2\theta})}=\\ \frac{\cos{\frac{4\cdot\color{red}{2\theta}}{2}}\cdot\sin{\frac{(4+1)\cdot\color{red}{2\theta}}{2}}}{\sin{\frac{\color{red}{2\theta}}{2}}}= \frac{\cos{(4\color{red}{\theta})}\cdot\sin{(5\color{red}{\theta}})}{\sin{\color{red}{\theta}}}$$

rtybase
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