Sines and cosines of angles in arithmetic progression

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

Solutions Collecting From Web of "Sines and cosines of angles in arithmetic progression"

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.

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