Question

What is $\sin \left( {{{\sin }^{ - 1}}\left( {\dfrac{1}{3}} \right)} \right)$?

Answer 1

Hint: To find the value of $\sin \left( {{{\sin }^{ - 1}}\left( {\dfrac{1}{3}} \right)} \right)$, first of all we will suppose the inverse function that is sine as x. Therefore, ${\sin ^{ - 1}}\dfrac{1}{3} = x$. Now, we will take the sine term to the RHS of the equation and get $\sin x = \dfrac{1}{3}$. Now, putting back the value of x, we will get the value of $\sin \left( {{{\sin }^{ - 1}}\left( {\dfrac{1}{3}} \right)} \right) = \dfrac{1}{3}$.

Complete step by step solution:
In this question, we have to solve $\sin \left( {{{\sin }^{ - 1}}\left( {\dfrac{1}{3}} \right)} \right)$. Here, we can see that there is an inverse function term in the given expression. Inverse trigonometric functions means the inverse functions of trigonometric functions. It is used to find the angle of the given trigonometric function.
According to definition, if
$\Rightarrow \sin A = x$
Then inverse of sine can be obtained by taking sine to the other side of the equal to sign.
$\Rightarrow A = {\sin ^{ - 1}}x$.
Note that, here A is the angle of the given function.
Here, we are going to solve our question by supposing the inverse function as x.
Let ${\sin ^{ - 1}}\dfrac{1}{3} = x$- - - - - - - - - - - (1)
Therefore, taking ${\sin ^{ - 1}}$ to the other side of the equal to sign, we get
$\Rightarrow \sin x = \dfrac{1}{3}$
Now, substituting the value of x from equation (1) in above equation, we get
$\Rightarrow \sin \left( {{{\sin }^{ - 1}}\dfrac{1}{3}} \right) = \dfrac{1}{3}$
Hence, this is our answer.

Note:
We can also find the value of $\sin \left( {{{\sin }^{ - 1}}\left( {\dfrac{1}{3}} \right)} \right)$ by solving the brackets in order.
First, calculate ${\sin ^{ - 1}}\dfrac{1}{3}$.
$\Rightarrow {\sin ^{ - 1}}\dfrac{1}{3} = 19.47122$
Now, we have to find the value of $\sin 19.4712206$.
$\Rightarrow \sin 19.4712206 = 0.33333$
Hence, $\sin \left( {{{\sin }^{ - 1}}\left( {\dfrac{1}{3}} \right)} \right) = 0.3333 = \dfrac{1}{3}$.
Note that this is a proven property of inverse trigonometric functions.
$\Rightarrow \sin \left( {{{\sin }^{ - 1}}x} \right) = x$