Question:
Solve for x :
$\cos \left(\sin ^{-1} x\right)=\frac{1}{2}$
Solution:
To find: value of x
Given: $\cos \left(\sin ^{-1} x\right)=\frac{1}{2}$
$\mathrm{LHS}=\cos \left(\sin ^{-1} \mathrm{x}\right)$
$=\cos \left(\cos ^{-1}\left(\sqrt{1-\mathrm{x}^{2}}\right)\right)$
$=\sqrt{1-\mathrm{x}^{2}}$
Therefore, $\sqrt{1-\mathrm{x}^{2}}=\frac{1}{2}$
Squaring both sides,
$1-x^{2}=\frac{1}{4}$
$x^{2}=1-\frac{1}{4}$
$x^{2}=\frac{3}{4}$
$x=\pm \frac{\sqrt{3}}{2}$
Therefore, $x=\pm \frac{\sqrt{3}}{2}$ are the required values of $x$.
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