# Sines and cosines of angles in arithmetic progression [closed]

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

$$\sum_{t=0}^{n}\cos{(\theta+t\phi)}$$

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

-

## closed as off-topic by Live Forever, Davide Giraudo, kjetil b halvorsen, Claude Leibovici, Jonas MeyerMay 5 '15 at 10:56

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Davide Giraudo, kjetil b halvorsen, Claude Leibovici
If this question can be reworded to fit the rules in the help center, please edit the question.

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
– Live Forever May 5 '15 at 0:15

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?

For

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

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.

-
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 $2\sin(\phi/2)$.

-