Taking into consideration the functions
$$\sum_{t=0}^{n} \sin{(\theta + t \phi)}=\frac{\sin({\frac{(n+1)\phi}2})\sin{(\theta+\frac{n \phi}2)}}{\sin{(\frac{\phi}2)}}$$
and
$$\sum_{t=0}^{n}\cos{(\theta+t\phi)}=\frac{\sin({\frac{(n+1)\phi}2})\cos{(\theta+\frac{n \phi}2)}}{\sin{(\frac{\phi}2)}}$$
Find, for all $\theta$, the values of
$$\sum_{t=0}^{n}\cos^{2}{(2t\theta)} \quad \textrm{and} \quad\sum_{t=0}^{n}\sin^{2}{(2t\theta)}$$
$$\sum_{r=0}^m\cos^2{(2r\alpha)}=\sum_{r=0}^m\frac{1+\cos4r\alpha}2=\frac {m+1}2+\frac12\sum_{r=0}^m\cos4r\alpha$$
$$\sum_{r=0}^m\sin^2{(2r\alpha)}=\sum_{r=0}^m\frac{1-\cos4r\alpha}2=\frac {m+1}2-\frac12\sum_{r=0}^m\cos4r\alpha$$
Now, $\sum_{r=0}^m\cos4r\alpha=\cos0+\cos4\alpha+\cos8\alpha+\cdots+\cos4m\alpha$
and $\sum_{t=0}^n\cos{(\theta+t\phi)}=\cos\theta+\cos(\theta+\phi)+\cdots+\cos(\theta+n\phi)$
So, comparing we get $\theta=0,\phi=4\alpha$ and $n=m$
