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Prove that if $\phi$ is not equal to $2k\pi$ for any integer $k$, then

$$\sum_{t=0}^{n} \sin{(\theta + t \phi)}=\frac{\sin({\frac{(n+1)\phi}2})\sin{(\theta+\frac{n \phi}2)}}{\sin{(\frac{\phi}2)}}$$

Find a similar formula for


where the functions sin and cos appear on the right-hand side.

Find, for all $\theta$, the values of

$$\sum_{t=0}^{n}\cos^{2}{(2t\theta)}$$ and $$\sum_{t=0}^{n}\sin^{2}{(2t\theta)}$$

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closed as off-topic by Bookend, Davide Giraudo, kjetil b halvorsen, Claude Leibovici, Jonas Meyer May 5 '15 at 10:56

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Put your equations in $...$ (inline) or $$...$$ (display) and write LaTeX. It looks much nicer. Wellcome! – vonbrand Feb 7 '13 at 19:57
@ElenaC: as an exercise: follow what I did (you can see by right-clicking on a rendered equation) and apply it to the rest of the post. – Ron Gordon Feb 7 '13 at 20:07
@rlgordonma thank you!! – ElenaC Feb 7 '13 at 21:16
up vote 1 down vote accepted

Use the exponential representation of the sines and cosines:

$$\cos{(\theta + t \phi)} = \frac{1}{2} \left ( e^{i (\theta + t \phi)} + e^{- (\theta + t \phi)} \right ) = \Re{[e^{i (\theta + t \phi)}]}$$

$$\sin{(\theta + t \phi)} = \frac{1}{2 i} \left ( e^{i (\theta + t \phi)} - e^{- (\theta + t \phi)} \right ) = \Im{[e^{i (\theta + t \phi)}]}$$

Then use a geometric series to sum.

Specifically, for the sine series, write

$$\begin{align}\sum_{t=0}^{n} \sin{(\theta + t \phi)} &= \Im{ \left [e^{i \theta} \sum_{t=0}^{n} e^{i t \phi} \right ]} \\ &= \Im{ \left [e^{i \theta} \frac{1-e^{i(n+1) \phi}}{1-e^{i\phi}} \right ]} \\ &=\Im{ \left [e^{i \theta} \frac{e^{i (n+1) \phi/2}}{e^{i \phi/2}} \frac{i 2 \sin{(n+1) \phi/2}}{i 2 \sin{\phi/2}} \right ]}\\ &= \Im{ \left [e^{i (\theta+n \frac{\phi}{2})} \right ]} \frac{\sin{\left [(n+1) \frac{\phi}{2} \right ]}}{\sin{\left (\frac{\phi}{2} \right )}}\\ &= \sin{ \left(\theta+n \frac{\phi}{2}\right)} \frac{\sin{\left [(n+1) \frac{\phi}{2} \right ]}}{\sin{\left (\frac{\phi}{2} \right )}}\\\end{align}$$

What is different for the cosine series?



write $\cos^{2}{(2t\theta)} = 1/2 + (1/2) \cos{(4 t \theta)}$ and see if the work you did for the cosine series applies. Similar for the $\sin^{2}{(2t\theta)}$ series.

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That, or write cosines as the real parts (and sines as the imaginary parts) of complex exponentials (sometimes simpler). – Did Feb 7 '13 at 20:19
@Did: that's how I always do them, but I chose to be a little more explicit at first. – Ron Gordon Feb 7 '13 at 20:22
@rlgordonma I was trying to do it by expanding the sum and letting it be equal to T, then multiply both sides by sin(phi/2) and use [cos(a-b)+cos(a+b)]=2sinasinb , but am stuck at that point. Is there a way to continue on this path? – ElenaC Feb 7 '13 at 21:23
I am not sure, but without seeing some steps done out, it is hard for me to see real progress being made. Were you able to follow my solution? – Ron Gordon Feb 7 '13 at 21:26
@rlgordonma yes, thank you so much! It definitely works that way. My concern is that, since I will handing in a write-up, I feel like the questions have been set up so that the answers are linked and flow into one another to give one final conclusion. So if I follow on with exponentials then I should be able to answer the whole question that way. I may be wrong though. – ElenaC Feb 7 '13 at 21:30

I'm adding another answer since there are people asking for solutions which do not use complex methods, so that this question can be used as a reference page. (Couldn't find such a thing already existing, please comment if there is.)

We have $$\eqalign{2\sin\Bigl(\frac\phi2\Bigr)\sum_{t=0}^n \sin(\theta+t\phi) &=\sum_{t=0}^n 2\sin(\theta+t\phi)\sin\Bigl(\frac\phi2\Bigr)\cr &=\sum_{t=0}^n \Bigl(\cos\bigl(\theta+(t-\tfrac12)\phi\bigr) -\cos\bigl(\theta+(t+\tfrac12)\phi\bigr)\Bigr)\cr &=\cos(\theta-\tfrac12\phi)-\cos(\theta+\tfrac12\phi)\cr &\qquad{}+\cos(\theta+\tfrac12\phi)-\cos(\theta+\tfrac32\phi)\cr &\qquad{}+\cos(\theta+\tfrac32\phi)-\cos(\theta+\tfrac52\phi)\cr &\qquad{}+\cdots\cr &\qquad{}+\cos(\theta+(n-\tfrac12)\phi)-\cos(\theta+(n+\tfrac12)\phi)\cr &=\cos(\theta-\tfrac12\phi)-\cos(\theta+(n+\tfrac12)\phi)\cr}$$ because all the intermediate terms cancel. Now use the identity $$\eqalign{\cos(\theta{}&{}-\tfrac12\phi)-\cos(\theta+(n+\tfrac12)\phi)\cr &=\cos\Bigl(\theta+\frac{n\phi}{2}-\frac{(n+1)\phi}{2}\Bigr) -\cos\Bigl(\theta+\frac{n\phi}{2}+\frac{(n+1)\phi}{2}\Bigr)\cr &=2\sin\Bigl(\theta+\frac{n\phi}{2}\Bigr)\sin\Bigl(\frac{(n+1)\phi}{2}\Bigr) \cr}$$ and then divide by $2\sin(\phi/2)$.

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