sin left{ {{sin }^{ - 1}}frac{1}{2} + {{cos } | vedclass

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(D) We know that for any $x \in [-1, 1]$,the identity ${{\sin }^{ - 1}}x + {{\cos }^{ - 1}}x = \frac{\pi }{2}$ holds true.Given the expression $\sin \

Vedclass · Accepted Answer

$1$

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(D) We know that for any $x \in [-1, 1]$,the identity ${{\sin }^{ - 1}}x + {{\cos }^{ - 1}}x = \frac{\pi }{2}$ holds true.Given the expression $\sin \left\{ {{\sin }^{ - 1}}\frac{1}{2} + {{\cos }^{ - 1}}\frac{1}{2} \right\}$.Substituting $x = \frac{1}{2}$ into the identity,we get:${{\sin }^{ - 1}}\frac{1}{2} + {{\cos }^{ - 1}}\frac{1}{2} = \frac{\pi }{2}$.Therefore,the expression becomes $\sin \left( \frac{\pi }{2} \right)$.Since $\sin \left( \frac{\pi }{2} \right) = 1$,the final answer is $1$.

$\sin \left\{ {{\sin }^{ - 1}}\frac{1}{2} + {{\cos }^{ - 1}}\frac{1}{2} \right\} = $

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