Find the simplified value of the following expressions ${sin}^{- 1} [cos {{sin}^{- 1} (\frac{\sqrt{3}}{2})}]$ .

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Solution

${sin}^{- 1} (\frac{\sqrt{3}}{2}) = θ$ such that $sin θ = \frac{\sqrt{3}}{2}$

$sin$ $θ = \frac{\sqrt{3}}{2} = sin$ $\frac{π}{3}$

$θ = \frac{π}{3}$

i.e. ${sin}^{- 1} (\frac{\sqrt{3}}{2}) = \frac{π}{3}$

Now Put $[{sin}^{- 1} (\frac{\sqrt{3}}{2}) = \frac{π}{3}]$ in given equation, we get

${sin}^{- 1} [cos (\frac{π}{3})]$

$⟹ {sin}^{- 1} (\frac{1}{2})$ Because $cos \frac{π}{3} = \frac{1}{2}$
Now,
${sin}^{- 1} (\frac{1}{2}) = θ$ such that $sin θ = \frac{1}{2}$

$sin$ $θ = \frac{1}{2} = sin$ $\frac{π}{6}$

$θ = \frac{π}{6}$

i.e. ${sin}^{- 1} (\frac{1}{2}) = \frac{π}{6}$

Therefore, ${sin}^{- 1} [cos {{sin}^{- 1} (\frac{\sqrt{3}}{2})}]$ $=$ $\frac{π}{6}$

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