Question

Write the value of cos $\left(2{\sin }^{-1}\frac{1}{3}\right)$.

Answer 1

Let $y={\sin }^{-1}\frac{1}{3}$
Then, $\sin y=\frac{1}{3}$

Now, $\cos y=\sqrt{1-{\sin }^{2}y}$,

$\Rightarrow \cos y=\sqrt{1-\frac{1}{9}}=\sqrt{\frac{8}{9}}=\frac{2\sqrt{2}}{3}$

$$\begin{gathered} \cos \left(2{\sin }^{-1}\frac{1}{3}\right)=\cos \left(2y\right) \\ ={\cos }^{2}y-{\sin }^{2}y\left[\because \cos 2x={\cos }^{2}x-{\sin }^{2}x\right] \\ ={\left(\frac{2\sqrt{2}}{3}\right)}^2-{\left(\frac{1}{3}\right)}^2 \\ =\frac{8}{9}-\frac{1}{9} \\ =\frac{7}{9} \end{gathered}$$

∴ $\cos \left(2{\sin }^{-1}\frac{1}{3}\right)=\frac{7}{9}$