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