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I have shown that $1+\cos(\theta)+\cos(2\theta)+\cos(3\theta)+\dots+\cos(n\theta)=\Re(\frac{1-e^{(n+1)i\theta}}{1-e^{i\theta}})=\frac{1-\cos(\theta)+\cos(n\theta)-\cos((n+1)\theta)}{2-2\cos(\theta)}=\frac{1}{2}+\frac{\cos(n\theta)-\cos((n+1)\theta)}{2-2\cos(\theta)}$

but I am unsure how to proceed from here. If possible, I would like to avoid using to many identities as this is an exercise in my complex analysis book.

A.B
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  • You will need to use the identity $cos(\theta)=cos^2(\frac{\theta}{2})-sin^2(\frac{\theta}{2})$ Also the $\frac{1}{2}$ term is immediate. If you want to avoid trig. identities, you need to use the exponential equivalents for sin and cos. – herb steinberg Feb 17 '20 at 02:19
  • @herbsteinberg Thank you. I'm still a bit confused on how to use that identity or how to get the result from converting $\cos(\theta)$ and $\sin(\theta)$ to $\frac{1}{2}(e^{i\theta}+e^{-i\theta})$ and $\frac{1}{2i}(e^{i\theta}-e^{-i\theta})$. Would you be able to expand on that? – A.B Feb 17 '20 at 02:55
  • I haven't tried to use the exponential identities. However as noted, there is a previous post for essentially the same problem. The basic idea for the numerator is $cos(n\theta)=cos((n+1/2)\theta)cos(\theta/2)+sin((n+1/2)\theta)sin(\theta/2)$ while $cos((n+1)\theta)=cos((n+1/2)\theta)cos(\theta/2)-sin((n+1/2)\theta)sin(\theta/2)$. Combine the two expressions to get the result. – herb steinberg Feb 17 '20 at 04:51

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