sum_(k=0)^90 cos^2(pi/(180)k) = ?

Oct 4, 2016

${\cos}^{2} \left(1\right) + \ldots + {\cos}^{2} \left({90}^{\circ}\right) = 45.5$

Explanation:

Remember that $\cos \left(\theta\right) = \sin \left({90}^{\circ} - \theta\right)$
and that $\cos \left({90}^{\circ}\right) = 0$

So
${\cos}^{2} \left(1\right) + \ldots . + {\cos}^{2} \left(89\right) + {\cos}^{2} \left(90\right)$
$\textcolor{w h i t e}{\text{XXX}} = {\cos}^{2} \left(1\right) + \ldots + {\cos}^{2} \left(89\right)$
$\textcolor{w h i t e}{\text{XXX}} = {\sin}^{2} \left(89\right) + \ldots + {\sin}^{2} \left(1\right)$
$\textcolor{w h i t e}{\text{XXX}} = {\sin}^{2} \left(1\right) + \ldots + {\sin}^{2} \left(89\right)$

If $s = {\cos}^{2} \left(1\right) + \ldots + {\cos}^{2} \left(89\right)$
then
$\textcolor{w h i t e}{\text{XXX}} 2 s = {\cos}^{2} \left(1\right) + \ldots + {\cos}^{2} \left(89\right)$
$\textcolor{w h i t e}{\text{XXXXX}} \underline{+ {\sin}^{2} \left(1\right) + \ldots + {\sin}^{2} \left(89\right)}$
$\textcolor{w h i t e}{\text{XXXXX")=underbrace(1color(white)("XXX")+...+color(white)("X")1)_("89 times}}$

$\rightarrow 2 s = 89$

$s = 44.5$

Oct 4, 2016

$44.5$

Explanation:

${\cos}^{2} \alpha = \frac{1}{2} \left(\cos \left(2 \alpha\right) + 1\right)$ so

${\sum}_{k = 1}^{n} {\cos}^{2} {\alpha}_{k} = \frac{n}{2} + \frac{1}{2} {\sum}_{k = 1}^{n} \cos \left(2 {\alpha}_{k}\right)$

now if ${\alpha}_{k} = \left(\frac{\pi}{2 n}\right) k$

${\sum}_{k = 1}^{n} \cos \left(2 {\alpha}_{k}\right) = \cos \left(\pi\right) = - 1$ (Cancellation by symmetry)

so

${\sum}_{k = 0}^{n} {\cos}^{2} \left(\frac{\pi}{2 n} k\right) = \frac{n}{2} - \frac{1}{2} = \frac{n - 1}{2}$

If $n = 90$ then

${\sum}_{k = 0}^{90} {\cos}^{2} \left(\frac{\pi}{180} k\right) = 44.5$