Question

Solve for x:

$\cos \left({\sin }^{-1}x\right)=\frac{1}{7}$

Answer 1

Given : $\cos \left({\sin }^{-1}x\right)=\frac{1}{7}$

Put ${\sin }^{-1}x=k$

$\Rightarrow x=\sin k$

Squaring both sides, we get;

$$\begin{gathered} x^2={\sin }^{2}k \\ \Rightarrow 1-x^2=1-{\sin }^{2}k \end{gathered}$$

$\Rightarrow 1-x^2={\cos }^{2}k....................\left(i\right)[Since,{\sin }^{2}x+{\cos }^{2}x=1]$

Now, $\cos k=\frac{1}{7}$ [Since,${\sin }^{-1}x=k$ and $\cos \left({\sin }^{-1}x\right)=\frac{1}{7}$ ]

Squaring both sides, we get;

${\cos }^{2}k=\frac{1}{49}......................\left(ii\right)$

From equation (i) and (ii)

$1-x^2=\frac{1}{49}$

$$\begin{gathered} \Rightarrow x^2=1-\frac{1}{49} \\ \Rightarrow x^2=\frac{49-1}{49} \end{gathered}$$

$\Rightarrow x=\sqrt{\frac{48}{49}}$

Thus the value of $x=\pm \frac{4\sqrt{3}}{7}$.