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$$\sum_{k=0}^n \cos(\alpha+k\theta)= \cos\left(\alpha+\left({n\theta\over2}\right)\right)*{\sin\left({(n+1)*\theta\over2}\right) \over \sin({\theta\over2})}$$

I have worked the left summation out to a couple of terms but I am not seeing how they are going to come together to get the right side. I wish I was better at writing on this site so I could show exactly what I have written. I also have to prove the summation of the sine but if I can figure out the cosine one above the sine should fall from that.

Sorry for the notation. I copied and pasted exactly like it said to do.

adjan
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ECollins
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    An interesting way to kill two birds with one stone is to look at the summation of $e^{i(\alpha+k\theta)}$, treating it as a geometric series, and then taking real and imaginary parts. – πr8 Jan 20 '16 at 14:17

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As remarked by πr8, the fastest way is to consider $\cos\phi$ as the real part of $e^{i\phi}$ transforming the problem into a geometric sum computation.

If you want to stay real, then multiply with the denominator and use that $$ 2\cos(α+kθ)\sin(\tfrac θ2)=\sin(α+(k+\tfrac12)θ)-\sin(α+(k-\tfrac12)θ) $$ to get a telescoping sum.

Lutz Lehmann
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