Simplify it.
$\sin \left( {2{{\sin }^{ - 1}}x} \right) = $?

Hint: In the above question, first we have to use an identity $\sin 2\theta = 2\sin \theta \cos \theta $ by considering $\theta = {\sin ^{ - 1}}x$. Then using the concept of inverse trigonometric function, we will find the value of $\sin \theta $ and then for $\cos \theta $ we will use the identity ${\sin ^2}\theta + {\cos ^2}\theta = 1$ so that we can find the value of $\cos \theta $ in terms of $\sin \theta $ whose value is known to us. In the end, we just have to substitute the values.

Complete step-by-step answer:
In the above question, we have to fins the value of $\sin \left( {2{{\sin }^{ - 1}}x} \right)$.
Here we have to use the formula $\sin 2\theta = 2\sin \theta \cos \theta $ .
Therefore, on comparing we get $\theta = {\sin ^{ - 1}}x$.
Therefore, we can write the above equation as
$\sin \left( {2{{\sin }^{ - 1}}x} \right) = 2\sin \left( {{{\sin }^{ - 1}}x} \right)\cos \left( {{{\sin }^{ - 1}}x} \right)$
Now, we can write as $\sin ({\sin ^{ - 1}}x) = x$ when $\theta \in \left( {\frac{\pi }{2}, - \frac{\pi }{2}} \right)$.
Therefore, $\sin ({\sin ^{ - 1}}x) = x..............\left( 1 \right)$
Also, we know that ${\sin ^2}\theta + {\cos ^2}\theta = 1$
Therefore, $\cos \theta = \sqrt {1 - {{\sin }^2}\theta } $
We can also write $\cos \left( {{{\sin }^{ - 1}}x} \right) = \cos \theta $ and $\theta = {\sin ^{ - 1}}x$
Therefore, $\cos \left( {{{\sin }^{ - 1}}x} \right) = \sqrt {1 - {x^2}} ...............\left( 2 \right)$
We have,
$\sin \left( {2{{\sin }^{ - 1}}x} \right) = 2\sin \left( {{{\sin }^{ - 1}}x} \right)\cos \left( {{{\sin }^{ - 1}}x} \right)$
Now substitute equation $\left( 1 \right)\,and\,\left( 2 \right)$ in above equation,
We get,
$\sin \left( {2{{\sin }^{ - 1}}x} \right) = 2x\sqrt {1 - {x^2}} $
Therefore, the required value is $2x\sqrt {1 - {x^2}} $.

Note: Inverse trigonometric functions are also called “Arc Functions” since, for a given value of trigonometric functions, they produce the length of arc needed to obtain that particular value. The inverse trigonometric functions perform the opposite operation of the trigonometric functions such as sine, cosine, tangent, cosecant, secant, and cotangent. We know that trigonometric functions are especially applicable to the right angle triangle.