Question

If $\cos [2{\sin }^{-1}(x\left)\right]=\frac{1}{9}$, then $x=?$

• A. only $\frac{2}{3}$
• B. only $-\frac{2}{3}$
• C. $\frac{2}{3},-\frac{2}{3}$
• D. Neither $\frac{2}{3}$ nor $-\frac{2}{3}$

Answer 1

The correct option is C

$\frac{2}{3},-\frac{2}{3}$

Explanation for the correct option:

Step 1. Find the value of $x$:

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

Let ${\sin }^{-1}\left(x\right)=A$

$\Rightarrow$ $\sin A=x$ ….(i)

Step 2. By Substituting ${\sin }^{-1}\left(x\right)=A$ in the given equation, we get

$\cos 2A=\frac{1}{9}$

$\Rightarrow$$1–2{\sin }^{2}A=\frac{1}{9}$

$\Rightarrow$ $2{\sin }^{2}A=1–\frac{1}{9}$

$\Rightarrow$ $2{\sin }^{2}A=\frac{8}{9}$

$\Rightarrow$ ${\sin }^{2}A=\frac{4}{9}$

$\Rightarrow$ $\begin{array}{rcl}\sin A& =& \sqrt{\frac{4}{9}}\end{array}$

$\begin{array}{rcl}& =& \pm \frac{2}{3}\end{array}$ ….(ii)

Step 3. By equating equation (i) and (ii), we get

$\mathit{x}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{2}}{\mathbf{3}}\mathbf{,}\mathbf{}\mathbf{-}\frac{\mathbf{2}}{\mathbf{3}}$

Hence, Option ‘C’ is Correct.