Logan Wyatt

2022-07-02

Compute $\sum _{k=0}^{180}{\mathrm{cos}}^{2}2{k}^{\circ }$

Keely Fernandez

Expert

Hint: Observe that for the first quadrant, we have
$\begin{array}{rl}& {\mathrm{cos}}^{2}{2}^{\circ }+{\mathrm{cos}}^{2}{4}^{\circ }+\cdots +{\mathrm{cos}}^{2}{88}^{\circ }+{\mathrm{cos}}^{2}{90}^{\circ }\\ & \phantom{\rule{1em}{0ex}}={\mathrm{cos}}^{2}{2}^{\circ }+{\mathrm{cos}}^{2}{4}^{\circ }+\cdots +{\mathrm{cos}}^{2}{44}^{\circ }+{\mathrm{sin}}^{2}{44}^{\circ }+{\mathrm{sin}}^{2}{42}^{\circ }+\cdots +{\mathrm{sin}}^{2}{2}^{\circ }+0\\ & \phantom{\rule{1em}{0ex}}=22.\end{array}$

Expert

$S=\sum _{r=0}^{180}{\mathrm{cos}}^{2}2{r}^{\circ }$
$2S=\sum _{r=0}^{180}\left(1+\mathrm{cos}4{r}^{\circ }\right)=181+\sum _{r=0}^{180}\mathrm{cos}4{r}^{\circ }$
Using $\sum \mathrm{cos}$ when angles are in arithmetic progression,
$\sum _{r=0}^{180}\mathrm{cos}4{r}^{\circ }=\frac{\mathrm{cos}\left(2\cdot {180}^{\circ }\right)\mathrm{sin}\left(2\cdot {181}^{\circ }\right)}{\mathrm{sin}{2}^{\circ }}=+1$

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