Compute $\sum_{k=0}^{180} \cos^22k^\circ $

Question

I'm trying to compute the following:

$$\sum_{k=0}^{180} \cos^22k^\circ $$

Answer 1

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}

Answer 2

$$\sum_{r=0}^{180}\cos^22r^\circ=1+\sum_{r=0}^{179}\cos^22r^\circ$$

$$S=\sum_{r=0}^{179}\cos^22r^\circ$$ $$=\sum_{r=0}^{44}\cos^22r^\circ+\sum_{r=45}^{89}\cos^22r^\circ+\sum_{r=90}^{134}\cos^22r^\circ+\sum_{r=135}^{179}\cos^22r^\circ$$

As $\cos(180^\circ\pm A)=-\cos A,\cos(360^\circ-A)=+\cos A,$

$$S=4\sum_{r=0}^{44}\cos^22r^\circ$$

Finally, $$\dfrac S2=2\sum_{r=0}^{44}\cos^22r^\circ=\sum_{r=0}^{44}\cos^22r^\circ+\sum_{r=0}^{44}\cos(90-2r)^\circ=45$$

Answer 3

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

Answer 4

$$S=\sum_{r=0}^{180}\cos^22r^\circ$$

$$2S=\sum_{r=0}^{180}(1+\cos4r^\circ)=181+\sum_{r=0}^{180}\cos4r^\circ$$

Using $\sum \cos$ when angles are in arithmetic progression,

$$\sum_{r=0}^{180}\cos4r^\circ=\dfrac{\cos(2\cdot180^\circ)\sin(2\cdot181^\circ)}{\sin2^\circ}=+1$$

Answer 5

$$S(n)=\cos^2 0°+ \cos^2 2°+ \cos^2 4°+....+ \cos^2 2n°$$

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 $$S_1=\frac{\sin\frac{nd}{2}\cdot \cos\frac{2a_1+d(n-1)}2}{\sin \frac d2}$$