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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

$1$

Vedclass · Answer

(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\} = $

Similar Questions

If $\cos ^{-1} 2x + \cos ^{-1} 3x = \frac{\pi}{3}$,then $x =$

$2{\tan ^{ - 1}}\left( {\frac{1}{3}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{7}} \right) = $

Let $f(x) = \cos^{-1}\left(\frac{2x}{1+x^2}\right) + \sin^{-1}\left(\frac{1-x^2}{1+x^2}\right)$,then the value of $f(1) + f(2)$ is -

If $y = \cos^{-1}\left( \frac{3\cos x + 4\sin x}{5} \right)$,then $\frac{dy}{dx} = $

Find the value of $\cos \left(\sec ^{-1} x+\csc ^{-1} x\right)$ for $|x| \geq 1$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module