Question

The value of ${\sin }^{-1}\cos \frac{29\pi }{5}$ is $\frac{a\pi }{b}$, where $a\&b$ are coprime. Then the value of $b^2-a^2$ is

Answer 1

${\sin }^{-1}\cos \frac{29\pi }{5}$ can be written as
$\frac{\pi }{2}-{\cos }^{-1}\cos (6\pi -\frac{\pi }{5})=\frac{\pi }{2}-\frac{\pi }{5}=\frac{3\pi }{10}$
Comparing with $\frac{a\pi }{b}$, we have $a=3$ and $b=10$
So $b^2-a^2=100-9=91$