Solve for $\mathbf{x}$ : $\cos \left(\sin ^{-1} x\right)=\frac{1}{9}$

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Solve for $\mathbf{x}$ : $\cos \left(\sin ^{-1} x\right)=\frac{1}{9}$

Solution:

To find: value of $x$
Given: $\cos \left(\sin ^{-1} x\right)=\frac{1}{9}$
$\mathrm{LHS}=\cos \left(\sin ^{-1} \mathrm{x}\right) \ldots(1)$
Let $\sin \theta=x$
Therefore $\theta=\sin ^{-1} \times \ldots(2)$
$\cos \theta=\sqrt{1-x^{2}}$
$\Rightarrow \theta=\cos ^{-1} \sqrt{1-x^{2}} \dots (3)$
From (2) and (3),
$\sin ^{-1} x=\cos ^{-1} \sqrt{1-x^{2}} \ldots \text { (4) }$
Substituting (4) in (1), we get
$\begin{array}{l} \text { LHS }=\cos \left(\cos ^{-1} \sqrt{1-x^{2}}\right) \\ =\sqrt{1-x^{2}} \end{array}$
Therefore, $\sqrt{1-x^{2}}=\frac{1}{9}$
Squaring and simplifying.
$\begin{array}{l} \Rightarrow 81-81 x^{2}=1 \\ \Rightarrow 81 x^{2}=80 \\ \Rightarrow x^{2}=\frac{80}{81} \\ \Rightarrow x=\pm \frac{4 \sqrt{5}}{9} \end{array}$
Therefore, $x=\pm \frac{4 \sqrt{5}}{9}$ are the required values of $x$ .