Question

Let $h(x)=\tan \left(\frac{\cos ^{-1}\left(\sin \left(\cos ^{-1} x\right)\right)+\sin ^{-1}\left(\cos \left(\sin ^{-1} x\right)\right)}{2}\right)$ then find the value of $\displaystyle\sum_{x=1}^7 h\left(\frac{x}{8}\right)$.

Answer 1

$\cos ^{-1}\left(\sin \left(\cos ^{-1} x\right)\right)+\sin ^{-1}\left(\cos \left(\sin ^{-1} x\right)\right)=\cos ^{-1} \sqrt{1-x^2}+\sin ^{-1} \sqrt{1-x^2}=\frac{\pi}{2}$
$\therefore h ( x )=1$
$\therefore \displaystyle\sum_{x=1}^7 h\left(\frac{x}{8}\right)=7$