Question

If $\cos \left(2{\sin }^{-1}x\right)=\frac{1}{9}$ then find $x$.

Answer 1

$\cos \left(2{\sin }^{-1}x\right)=\frac{1}{9}$

We know $\cos 2\theta =1-2{\sin }^2\theta$

We can apply this formula have so

$1-{\sin }^2\left({\sin }^{-1}x\right)=\frac{1}{9}$

$2\left[\sin \right({\sin }^{-1}x){]}^2=1-\frac{1}{9}$

We know that $\sin \left({\sin }^{-1}x\right)=x$ so

$2x^2=\frac{8}{9}$

$x^2=\frac{4}{9}$

$x=\pm \frac{2}{3}$

$x=+\frac{2}{3},x=-\frac{2}{3}$