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

Hint: In the problem we have an inverse trigonometric function and a normal trigonometric function. So, we will first solve the inverse trigonometric function, for this we will assume the inverse trigonometric ratio to a variable let us say $x$. Now we will simplify and calculate the value of the inverse trigonometric function. From this value we will calculate the value of the given function by substituting the calculated inverse trigonometric function.

Complete step-by-step answer:
Given that, $\sin \left( {{\cos }^{-1}}\left( \dfrac{\sqrt{2}}{2} \right) \right)$
In the above equation we have inverse trigonometric function ${{\cos }^{-1}}$ and the normal trigonometric function $\sin $. Assuming the inverse trigonometric function as a variable let us say $x$. Then we will have
${{\cos }^{-1}}\left( \dfrac{\sqrt{2}}{2} \right)=x$
Applying the $\cos $ function on the both sides of the above equation. Then we will get
$\Rightarrow \cos \left( {{\cos }^{-1}}\left( \dfrac{\sqrt{2}}{2} \right) \right)=\cos x$
We know that a function will be cancelled when it is multiplied by its inverse function, then we will have
$\Rightarrow \dfrac{\sqrt{2}}{2}=\cos x$
We have the $\cos \left( \dfrac{\pi }{4} \right)=\dfrac{\sqrt{2}}{2}$, substituting this value in the above equation, then we will get
$\Rightarrow \cos \left( \dfrac{\pi }{4} \right)=\cos x$
Applying inverse function on both sides of the above equation and cancelling the terms that are possible in the above equation, then we will get
$\Rightarrow x=\dfrac{\pi }{4}$
Hence the value of ${{\cos }^{-1}}\left( \dfrac{\sqrt{2}}{2} \right)$ will be
${{\cos }^{-1}}\left( \dfrac{\sqrt{2}}{2} \right)=\dfrac{\pi }{4}$
Substituting this value in the given equation, then we will get
$\Rightarrow \sin \left( {{\cos }^{-1}}\left( \dfrac{\sqrt{2}}{2} \right) \right)=\sin \left( \dfrac{\pi }{4} \right)$
We have the value of $\sin \left( \dfrac{\pi }{4} \right)=\dfrac{\sqrt{2}}{2}$.
Hence $\sin \left( {{\cos }^{-1}}\left( \dfrac{\sqrt{2}}{2} \right) \right)=\dfrac{\sqrt{2}}{2}$.

Note: In this problem we have assumed the inverse trigonometric function to a variable for our convenience. If you are familiar with the values of inverse trigonometric functions, then you can simplify the use of that value. But it is not preferable to go with this method. So, follow the above method to evaluate any kind of equation for different angles.