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In working to prove that $$1+\cos\theta+\cos(2\theta)+\dots+\cos(n\theta)=\frac{1}{2}+\frac{\sin((n+\frac{1}{2})\theta)}{2\sin(\theta/2)} \tag{1}$$

I have shown $$\begin{align} 1+\cos(\theta)+\cos(2\theta)+\cos(3\theta)+\dots+\cos(n\theta) &=\Re\left(\frac{1-e^{(n+1)i\theta}}{1-e^{i\theta}}\right) \\[6pt] &=\frac{1-\cos(\theta)+\cos(n\theta)-\cos((n+1)\theta)}{2-2\cos(\theta)} \\[6pt] &=\frac{1}{2}+\frac{\cos(n\theta)-\cos((n+1)\theta)}{2-2\cos(\theta)} \tag{2} \end{align}$$ but I am unsure how to proceed from here and get the last term of $(2)$ to match the last term of $(1)$:

$$\frac{\cos(n\theta)-\cos((n+1)\theta)}{2-2\cos(\theta)}=\frac{\sin((n+\frac{1}{2})\theta)}{2\sin(\theta/2)} \tag{3}$$

I have read this post "How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression?", but I cannot seem to convert from the form their answer is in to my form.

If possible, I would like to avoid using too many identities as this is an exercise in my complex analysis book.

Blue
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A.B
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    From where you are, simply apply the difference-to-product identity $$\cos A - \cos B = -2\sin\frac12(A+B)\sin\frac12(A-B)$$ and the half-angle identity $$1-\cos A= 2\sin^2\frac12A$$ I hope that isn't "too many identities". :) – Blue Feb 17 '20 at 03:09
  • I have edited your question to highlight the specific aspect of the problem that has you stumped. If this doesn't meet with your approval, feel free to re-edit or simply roll-back the changes. – Blue Feb 17 '20 at 03:22
  • @Blue Thanks so much! The first identity I'm not familiar with, but the second one I am. Do you know if there might be another way to reach the conclusion with different identities? I am familiar with double/half angle identities, angle sum/difference identities, and the Pythagorean identities. I think I could also finish it by using the exponential form of $\sin\theta$ and $\cos\theta$, but that might be more complicated. – A.B Feb 17 '20 at 03:24
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    As for the first identity I mentioned, it might be helpful to view it as $$\cos(\theta+\phi)-\cos(\theta-\phi) = -2\sin\theta\sin\phi$$ where the $\cos\theta\cos\phi$ terms of the expanded form have canceled and the $-\sin\theta\sin\phi$ terms have combined. That identity is pretty-much the go-to result to get to the specific form you are trying to achieve. Of course, I think my answer to the question you linked show a pretty good alternative approach. :) – Blue Feb 17 '20 at 03:31

2 Answers2

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To prove

$$\frac{\cos(n\theta)-\cos((n+1)\theta)}{2-2\cos(\theta)}=\frac{\sin((n+\frac{1}{2})\theta)}{2\sin(\theta/2)}$$

examine the LHS

$$LHS = \frac{\cos(n\theta)-\cos(n\theta)\cos\theta+\sin(n\theta)\sin\theta}{2-2\cos\theta}$$ $$= \frac{\cos(n\theta)(1-\cos\theta)+\sin(n\theta)\sin\theta}{2(1-\cos\theta)}$$

Then, use $1-\cos\theta=2\sin^2\frac\theta2$ and $\sin\theta = 2\sin\frac\theta2 \sin\frac\theta2$ to simplify,

$$LHS= \frac{\cos(n\theta)\sin\frac\theta2+\sin(n\theta)\cos\frac\theta2}{2\sin\frac\theta2}=\frac{\sin((n+\frac{1}{2})\theta)}{2\sin\frac\theta2}=RHS$$

Quanto
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$1+\cos(\theta)+\cos(2\theta)+\cos(3\theta)+\dots+\cos(n\theta)=$

$\dfrac{2+e^{i\theta}+e^{-i\theta}+e^{2i\theta}+e^{-2i\theta}+e^{3i\theta}+e^{-3i\theta}\dots+e^{ni\theta}+e^{-ni\theta}}2=$

$\dfrac{1+e^{-ni\theta}+\dots+e^{-3i\theta}+e^{-2i\theta}+e^{-i\theta}+1+e^{i\theta}+e^{2i\theta}+e^{3i\theta}+\cdots+e^{ni\theta}}2=$

$\dfrac12$+$\dfrac{e^{-ni\theta}\left(\dfrac{1-e^{(2n+1)i\theta}}{1-e^{i\theta}}\right)}2=$

$\dfrac12$+$\dfrac{e^{-ni\theta}\left(\dfrac{e^{-i\theta/2}-e^{(2n+1/2)i\theta}}{e^{-i\theta/2}-e^{i\theta/2}}\right)}2=$

$\dfrac12$+$\dfrac{ \left(\dfrac{e^{-i(n+1/2)\theta}-e^{i(n+1/2)\theta}}{e^{-i\theta/2}-e^{i\theta/2}}\right)}2=$

$\dfrac12+\dfrac12\dfrac{ \sin\left((n+\frac12)\theta\right)}{\sin(\theta/2)}$

J. W. Tanner
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