Q10:

Find the values of each of the expressions: $\sin^{-1}(\sin(\frac{2\pi}{3}))$

Step 1: Analyzing the Domain and Range

The principal value branch of the inverse sine function, $\sin^{-1}(x)$, is defined on the closed interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$. [Per the definition of principal value branches of inverse trigonometric functions].

Given the expression $\sin^{-1}(\sin(\frac{2\pi}{3}))$, we observe that the argument $\frac{2\pi}{3}$ lies outside the principal interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$ because $\frac{2\pi}{3} \approx 2.094$ radians and $\frac{\pi}{2} \approx 1.571$ radians. Since $\frac{2\pi}{3} > \frac{\pi}{2}$, we cannot directly apply the property $\sin^{-1}(\sin \theta) = \theta$.

Step 2: Utilizing Trigonometric Identity

We use the supplementary angle property of the sine function: $\sin(\pi - \theta) = \sin(\theta)$. [Per the Allied Angle Theorems].

Applying this to the angle $\frac{2\pi}{3}$:

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

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

Step 3: Evaluating the Inverse Function

Substitute the simplified value back into the original expression:

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

Since $\frac{\pi}{3} \in [-\frac{\pi}{2}, \frac{\pi}{2}]$, the condition for the identity $\sin^{-1}(\sin x) = x$ is now satisfied. [By the Principle of Inverse Functions].

$\sin^{-1}\left(\sin\left(\frac{\pi}{3}\right)\right) = \frac{\pi}{3}$

Step 4: Geometric Representation of the Argument Mapping

The SVG above demonstrates how the angle $\frac{2\pi}{3}$ (in Quadrant II) shares the same vertical projection as the angle $\frac{\pi}{3}$ (in Quadrant I) on the unit circle, confirming the equality of their sine values.

Final Solution: \frac{\pi}{3}