Find the value of sin⁻¹(sin (4π/3)) is

The correct option is -(π/3).

To find the value of ${\sin }^{-1}\left(\sin \left(\frac{4\pi }{3}\right)\right)$, we need to understand the definition and range of the principal value branch of the inverse sine function.

The principal value branch of ${\sin }^{-1}\left(x\right)$ is restricted to the interval:
$\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$

This means that ${\sin }^{-1}\left(\sin \right(\theta \left)\right)=\theta$ if and only if $\theta$ lies within this interval.
Here, the given angle is $\theta =\frac{4\pi }{3}$. Since $\frac{4\pi }{3}$ is greater than $\frac{\pi }{2}$, it does not lie in the principal value interval $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$. Therefore, we cannot write ${\sin }^{-1}\left(\sin \left(\frac{4\pi }{3}\right)\right)=\frac{4\pi }{3}$ directly.

We must rewrite the angle $\frac{4\pi }{3}$ in a form that corresponds to an angle within the principal value interval while preserving the sine value.
We can express $\frac{4\pi }{3}$ as:
$\frac{4\pi }{3}=\pi +\frac{\pi }{3}$

Using the trigonometric identity $\sin (\pi +x)=-\sin \left(x\right)$, we have:
$\sin \left(\frac{4\pi }{3}\right)=\sin \left(\pi +\frac{\pi }{3}\right)=-\sin \left(\frac{\pi }{3}\right)$

Now, we use another identity, $-\sin \left(x\right)=\sin (-x)$:
$-\sin \left(\frac{\pi }{3}\right)=\sin \left(-\frac{\pi }{3}\right)$

Substituting this back into our original expression, we get:
${\sin }^{-1}\left(\sin \left(\frac{4\pi }{3}\right)\right)={\sin }^{-1}\left(\sin \left(-\frac{\pi }{3}\right)\right)$

Since the angle $-\frac{\pi }{3}$ lies within the principal value interval $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$, we can apply the property ${\sin }^{-1}\left(\sin \right(\theta \left)\right)=\theta$:
${\sin }^{-1}\left(\sin \left(-\frac{\pi }{3}\right)\right)=-\frac{\pi }{3}$

Thus, the value of the expression is $-\frac{\pi }{3}$.