Question

Let $p =\displaystyle\sum_{ n =1}^{9} \sin ^{2}\left(\frac{ n \pi}{24}\right)$ and $q =\displaystyle\sum_{ n =1}^{9} \cos ^{2}\left(\frac{ n \pi}{24}\right)$, then find the value of $(p+q)$.

Answer 1

$p + q =\displaystyle\sum_{ n =1}^{9} \sin ^{2}\left(\frac{ n \pi}{24}\right)+\sum_{ n =1}^{9} \cos ^{2}\left(\frac{ n \pi}{24}\right)$
$=\displaystyle\sum_{ n =1}^{9}\left[\sin ^{2}\left(\frac{ n \pi}{24}\right)+\cos ^{2}\left(\frac{ n \pi}{24}\right)\right]=\sum_{ n =1}^{9} 1=9$