Find the value of \(\cos \left( {{{\sin }^{ - 1}}\frac{1}{5}} \right)\

1.	Find the value of \(\cos \left( {{{\sin }^{ - 1}}\frac{1}{5}} \right)\).1. \(\frac{1}{{\sqrt 5 }}\)2. \(\frac{{2\sqrt 6 }}{5}\)3. \(- \frac{1}{{\sqrt 5 }}\)4. \( - \frac{{2\sqrt 6 }}{5}\)
Answer» Correct Answer - Option 2 : \(\frac{{2\sqrt 6 }}{5}\) Concept: \({\cos ^{ - 1}}\left( {\cos x} \right) = x\) Calculation: \(\cos \left( {{{\sin }^{ - 1}}\frac{1}{5}} \right)\) Let, \({\sin ^{ - 1}}\frac{1}{5} = x\) \(\sin x = \frac{1}{5}\) We know, \({\sin ^2}x + {\cos ^2}x = 1\) \({\left( {\frac{1}{5}} \right)^2} + {\cos ^2}x = 1\) \({\cos ^2}x = 1 - {\left( {\frac{1}{5}} \right)^2}\) \({\cos ^2}x = \frac{{24}}{{25}}\) \(\cos x = \frac{{2\sqrt 6 }}{5}\) \(x = {\cos ^{ - 1}}\frac{{2\sqrt 6 }}{5}\) Now substitute for \({\sin ^{ - 1}}\frac{1}{5}\) in \(\cos \left( {{{\sin }^{ - 1}}\frac{1}{5}} \right)\) \(\cos \left( {{{\sin }^{ - 1}}\frac{1}{5}} \right) = \cos \left( {{{\cos }^{ - 1}}\frac{{2\sqrt 6 }}{5}} \right)\) \( = \frac{{2\sqrt 6 }}{5}\)

Find the value of \(\cos \left( {{{\sin }^{ - 1}}\frac{1}{5}} \right)\).1. \(\frac{1}{{\sqrt 5 }}\)2. \(\frac{{2\sqrt 6 }}{5}\)3. \(- \frac{1}{{\sqrt 5 }}\)4. \( - \frac{{2\sqrt 6 }}{5}\)

Answer» Correct Answer - Option 2 : \(\frac{{2\sqrt 6 }}{5}\)

Concept:

\({\cos ^{ - 1}}\left( {\cos x} \right) = x\)

Calculation:

\(\cos \left( {{{\sin }^{ - 1}}\frac{1}{5}} \right)\)

Let,

\({\sin ^{ - 1}}\frac{1}{5} = x\)

\(\sin x = \frac{1}{5}\)

We know,

\({\sin ^2}x + {\cos ^2}x = 1\)

\({\left( {\frac{1}{5}} \right)^2} + {\cos ^2}x = 1\)

\({\cos ^2}x = 1 - {\left( {\frac{1}{5}} \right)^2}\)

\({\cos ^2}x = \frac{{24}}{{25}}\)

\(\cos x = \frac{{2\sqrt 6 }}{5}\)

\(x = {\cos ^{ - 1}}\frac{{2\sqrt 6 }}{5}\)

Now substitute for \({\sin ^{ - 1}}\frac{1}{5}\) in \(\cos \left( {{{\sin }^{ - 1}}\frac{1}{5}} \right)\)

\(\cos \left( {{{\sin }^{ - 1}}\frac{1}{5}} \right) = \cos \left( {{{\cos }^{ - 1}}\frac{{2\sqrt 6 }}{5}} \right)\)

\( = \frac{{2\sqrt 6 }}{5}\)