Inverse Trigonometric Functions

To determine the principal value of $\sin^{-1}\left(\sin\left(\frac{2\pi}{3}\right)\right)$, we need to understand the properties of the inverse sine function, also known as the arcsine function.

The sine function, $\sin(x)$, is periodic with period $2\pi$, meaning that for any integer $k$, we have:

$\sin(x) = \sin(x + 2k\pi)$.

The inverse sine function, $\sin^{-1}(x)$, has a restricted domain, typically taken as:

$-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$,

where $y = \sin^{-1}(x)$.

Given the expression inside the inverse sine function:

$\sin\left(\frac{2\pi}{3}\right)$,

we know:

$\sin\left(\frac{2\pi}{3}\right) = \sin\left(\pi - \frac{\pi}{3}\right) = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$.

Now, we want to find the principal value of $\sin^{-1}\left(\sin\left(\frac{2\pi}{3}\right)\right)$.

Let:

$x = \frac{2\pi}{3}$,

so:

$\sin\left(\frac{2\pi}{3}\right) = \sin(x) = \frac{\sqrt{3}}{2}$.

To find the principal value, we need to find an angle $y$ within the range $-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$ such that $\sin(y) = \frac{\sqrt{3}}{2}$.

The principal value that satisfies this condition is:

$y = \sin^{-1}\left(\frac{\sqrt{3}}{2}\right)$.

Given the range of the inverse sine function, the corresponding angle is:

$y = \frac{\pi}{3}$.

Now, we need to match this value with one of the given options. Since there is no option for $\frac{\pi}{3}$ directly, but recalling the properties of the sine function and the range, and knowing that $\frac{2\pi}{3}$ is not within the range of the principal values, we need to reconsider equivalent values.

None of the given options match exactly the principal value we calculated, which is $\frac{\pi}{3}$. Therefore, the correct answer is:

Option D: none