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