Question

If cos(2sin^-1x) = $\frac{1}{9}$, then the value of x is ______________.

Answer 1

Let ${\sin }^{-1}x=\theta \Rightarrow \sin \theta =x$.

$$\begin{gathered} \therefore \cos \left(2{\sin }^{-1}x\right)=\frac{1}{9} \\ \Rightarrow \cos 2\theta =\frac{1}{9} \\ \Rightarrow 1-2{\sin }^2\theta =\frac{1}{9} \end{gathered}$$
$$\begin{gathered} \Rightarrow 2{\sin }^2\theta =1-\frac{1}{9}=\frac{8}{9} \\ \Rightarrow {\sin }^2\theta =\frac{4}{9} \\ \Rightarrow \sin \theta =\pm \frac{2}{3} \\ \Rightarrow x=\pm \frac{2}{3}\left(\sin \theta =x\right) \end{gathered}$$

Thus, the value of x is $\pm \frac{2}{3}$.

If cos(2sin⁻¹x) = $\frac{1}{9}$, then the value of x is $\underline{\pm \frac{2}{3}}$.