Compute $\sum_{k=0}^{180} \cos^22k^\circ $ I'm trying to compute the following:
$$\sum_{k=0}^{180} \cos^22k^\circ $$
 A: Hint: 
Observe that for the first quadrant, we have
\begin{align}
&\cos^22^\circ+\cos^24^\circ+\cdots+\cos^288^\circ+\cos^290^\circ\\
&\quad=\cos^22^\circ+\cos^24^\circ+\cdots+\cos^244^\circ
+\sin^244^\circ+\sin^242^\circ+\cdots+\sin^22^\circ+0\\
&\quad=22.
\end{align}
A: $$\sum_{r=0}^{180}\cos^22r^\circ=1+\sum_{r=0}^{179}\cos^22r^\circ$$
$$S=\sum_{r=0}^{179}\cos^22r^\circ$$
$$=\sum_{r=0}^{44}\cos^22r^\circ+\sum_{r=45}^{89}\cos^22r^\circ+\sum_{r=90}^{134}\cos^22r^\circ+\sum_{r=135}^{179}\cos^22r^\circ$$
As $\cos(180^\circ\pm A)=-\cos A,\cos(360^\circ-A)=+\cos A,$
$$S=4\sum_{r=0}^{44}\cos^22r^\circ$$
Finally,
$$\dfrac S2=2\sum_{r=0}^{44}\cos^22r^\circ=\sum_{r=0}^{44}\cos^22r^\circ+\sum_{r=0}^{44}\cos(90-2r)^\circ=45$$
A: Remark that:
$cos(x+90°) = sin(x)$
And that:
$cos^2(x) + sin^2(x) = 1$
So you can group your terms:
$$
\begin{align*}
\sum_{x=0}^{180} cos^2(x.2°) &=  \sum_{x=0}^{44} cos^2(x.2°) \\
                             & + \sum_{x=45}^{89} cos^2(x.2°) \\
                             & + \sum_{x=90}^{134} cos^2(x.2°) \\
                             & + \sum_{x=135}^{179}{cos^2(x.2°)} \\
                             & + cos^2(180.2°) \\
\\
&= \sum_{x=0}^{44}{cos^2(x.2°)+cos^2(x.2°+90°)} \\
& + \sum_{x=90}^{179}{cos^2(x.2°)+cos^2(x.2°+90°)} \\
& + cos^2(360°) \\
\\
&= \sum_{x=0}^{44}{cos^2(x.2°)+sin^2(x.2°)} \\
& + \sum_{x=90}^{134}{cos^2(x.2°)+sin^2(x.2°)} \\
& + 1 \\
\\
&= \sum_{x=0}^{44}{1} \\
& + \sum_{x=90}^{134}{1} \\
& + 1 \\
&= 91
\end{align*}
$$
See the approximate results at wolfram alpha
A: $$S=\sum_{r=0}^{180}\cos^22r^\circ$$
$$2S=\sum_{r=0}^{180}(1+\cos4r^\circ)=181+\sum_{r=0}^{180}\cos4r^\circ$$
Using $\sum \cos$ when angles are in arithmetic progression,
$$\sum_{r=0}^{180}\cos4r^\circ=\dfrac{\cos(2\cdot180^\circ)\sin(2\cdot181^\circ)}{\sin2^\circ}=+1$$
A: $$S(n)=\cos^2 0°+ \cos^2 2°+ \cos^2 4°+....+ \cos^2 2n°$$
1) $\cos^2 \alpha =\frac{1+\cos 2 \alpha}2$. Then 
$$S=\frac{1+n+\left(\cos 4°+ \cos 8°+....+ \cos 4n°\right)}2$$
2) Let $S_1=\cos a_1+ \cos a_2+....+ \cos a_n$, where $(a_n) -$ arithmetical progression ($d\not=2\pi k, k \in \mathbb Z$)
Then 
$$S_1=\frac{\sin\frac{nd}{2}\cdot \cos\frac{2a_1+d(n-1)}2}{\sin \frac d2}$$
