Question

If $({\sin }^{-1}x{)}^2+({\cos }^{-1}x{)}^2=\frac{17{\pi }^2}{36}$, find $x$.

Answer 1

Let,

$\frac{17{\pi }^2}{36}=(si{n}^{-1}x+co{s}^{-1}x{)}^2-2si{n}^{-1}xco{s}^{-1}x$ ($a^2+b^2=(a+b{)}^2-2ab$)

$\frac{17{\pi }^2}{36}=\frac{{\pi }^2}{4}-2si{n}^{-1}[\frac{\pi }{2}-si{n}^{-1}x]$

$(si{n}^{-1}{)}^2-\frac{\pi }{2}(si{n}^{-1}x)-\frac{{\pi }^2}{9}=0$

$si{n}^{-1}x=\frac{\frac{\pi }{2}\pm \sqrt{\frac{{\pi }^2}{4}+\frac{4{\pi }^2}{9}}}{2\left(1\right)}$

$si{n}^{-1}x=\frac{\pi }{2},-\frac{\pi }{6}$

$si{n}^{-1}x=-\frac{\pi }{6}$

$x=-\frac{1}{2}$