How do you evaluate $\sin \left[ {2{{\cos }^{ - 1}}\left( {\dfrac{3}{5}} \right)} \right]$?

Hint: First apply the formula of sine, $\sin 2x = 2\sin x\cos x$. Then, cancel out cosine inverse with cosine. After that convert cosine inverse into sine inverse by the formula ${\sin ^2}x + {\cos ^2}x = 1$. Also, cancel out sine with sine inverse. After that, multiply the terms to get the desired result.

Complete step-by-step solution:
The given expression is $\sin \left[ {2{{\cos }^{ - 1}}\left( {\dfrac{3}{5}} \right)} \right]$.
We know that,
$\sin 2x = 2\sin x\cos x$
Apply the above identity in the expression,
$ \Rightarrow \sin \left[ {2{{\cos }^{ - 1}}\left( {\dfrac{3}{5}} \right)} \right] = 2\sin \left[ {{{\cos }^{ - 1}}\left( {\dfrac{3}{5}} \right)} \right]\cos \left[ {{{\cos }^{ - 1}}\left( {\dfrac{3}{5}} \right)} \right]$
Cancel out cosine with cosine inverse,
$ \Rightarrow \sin \left[ {2{{\cos }^{ - 1}}\left( {\dfrac{3}{5}} \right)} \right] = 2\sin \left[ {{{\cos }^{ - 1}}\left( {\dfrac{3}{5}} \right)} \right] \times \dfrac{3}{5}$
Multiply the terms,
$ \Rightarrow \sin \left[ {2{{\cos }^{ - 1}}\left( {\dfrac{3}{5}} \right)} \right] = \dfrac{6}{5}\sin \left[ {{{\cos }^{ - 1}}\left( {\dfrac{3}{5}} \right)} \right]$...............….. (1)
Now let us assume $x = {\cos ^{ - 1}}\left( {\dfrac{3}{5}} \right)$.
Take cos on both sides,
$ \Rightarrow \cos x = \cos \left( {{{\cos }^{ - 1}}\left( {\dfrac{3}{5}} \right)} \right)$
Cancel out cosine with cosine inverse,
$ \Rightarrow \cos x = \dfrac{3}{5}$
We know that,
${\sin ^2}x + {\cos ^2}x = 1$
Substitute the value in the above formula,
$ \Rightarrow {\sin ^2}x + {\left( {\dfrac{3}{5}} \right)^2} = 1$
Square the term,
$ \Rightarrow {\sin ^2}x + \dfrac{9}{{25}} = 1$
Move constant part on the right side,
$ \Rightarrow {\sin ^2}x = 1 - \dfrac{9}{{25}}$
Take LCM on the right side,
$ \Rightarrow {\sin ^2}x = \dfrac{{25 - 9}}{{25}}$
Simplify the term,
$ \Rightarrow {\sin ^2}x = \dfrac{{16}}{{25}}$
Take the square root on both sides,
$ \Rightarrow \sin x = \dfrac{4}{5}$
Take ${\sin ^{ - 1}}$ on both sides,
$ \Rightarrow {\sin ^{ - 1}}\left( {\sin x} \right) = {\sin ^{ - 1}}\dfrac{4}{5}$
Cancel out sine inverse with sine,
$ \Rightarrow x = {\sin ^{ - 1}}\dfrac{4}{5}$
Then, we can say that,
$ \Rightarrow {\cos ^{ - 1}}\dfrac{3}{5} = {\sin ^{ - 1}}\dfrac{4}{5}$
Substitute the values in equation (1),
\[ \Rightarrow \sin \left[ {2{{\cos }^{ - 1}}\left( {\dfrac{3}{5}} \right)} \right] = \dfrac{6}{5}\sin \left[ {{{\sin }^{ - 1}}\left( {\dfrac{4}{5}} \right)} \right]\]
Cancel out sine inverse with sine,
\[ \Rightarrow \sin \left[ {2{{\cos }^{ - 1}}\left( {\dfrac{3}{5}} \right)} \right] = \dfrac{6}{5} \times \dfrac{4}{5}\]
Multiply the terms,
\[ \Rightarrow \sin \left[ {2{{\cos }^{ - 1}}\left( {\dfrac{3}{5}} \right)} \right] = \dfrac{{24}}{{25}}\]

Hence, the value of \[\sin \left[ {2{{\cos }^{ - 1}}\left( {\dfrac{3}{5}} \right)} \right]\] is \[\dfrac{{24}}{{25}}\].

Note: Inverse trigonometry formulas can help you solve any related questions. Just as addition is an inverse of subtraction and multiplication is an inverse of division, in the same way, inverse functions in an inverse trigonometric function. We can call it by different names such as anti-trigonometric functions, arcus functions, and cyclometric functions. The inverse trigonometric functions are as popular as anti-trigonometric functions. The inverse functions have the same name as functions but with a prefix “arc” so the inverse of sine will be arcsine, the inverse of cosine will be arccosine, and tangent will be arctangent.