Q.Find the principal value of the following: $\sin^{-1}\left(\sin\frac{2\pi}{3}\right)$

The key idea is that $\sin^{-1}$ returns the principal value in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, not the original angle if it lies outside this range.

1. The given angle is $\frac{2\pi}{3}$, which is in Quadrant II. Its sine is positive: $\sin\frac{2\pi}{3} = \sin\left(\pi - \frac{\pi}{3}\right) = \sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}$. …

The function $\sin^{-1}$ (also written as $\arcsin$) is the inverse of the sine function, but only when sine is restricted to a specific interval. Without that restriction, sine is not one-to-one — many different angles give the same sine value. So $\sin^{-1}$ is defined to return only the principal value: the unique angle in the interval $[-\pi/2, \pi/2]$ whose sine equals the given number.

When you see $\sin^{-1}(\sin \theta)$, the instinct might be to cancel and say $\theta$. But that only works if $\theta$ itself lies in $[-\pi/2, \pi/2]$. If $\theta$ is outside that range, you must find the angle inside the range that has the same sine.

Here, $\theta = \frac{2\pi}{3}$. Let's check: $\frac{2\pi}{3} = 120^\circ$, which is well outside $[-\pi/2, \pi/2] = [-90^\circ, 90^\circ]$. So we cannot simply cancel.

1. Find the sine of the given angle.

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

   This uses the identity $\sin(\pi - x) = \sin x$.

2. Now ask: what angle in $[-\pi/2, \pi/2]$ has sine $\frac{\sqrt{3}}{2}$?

   The standard angle is $\frac{\pi}{3}$ (60°), and $\frac{\pi}{3}$ is indeed in $[-\pi/2, \pi/2]$.

   No other angle in that interval gives the same sine — $\sin x$ is one-to-one there.

3. Therefore: …

Method: Evaluating $\sin^{-1}(\sin\theta)$

Use this whenever you must simplify $\sin^{-1}(\sin\theta)$ for an angle $\theta$ that may lie outside the principal range of $\sin^{-1}$.

Steps

Step 1: Recall the principal range and test $\theta$.

$\sin^{-1}$ always returns an angle in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. If $\theta$ already lies in this interval, the answer is simply $\theta$ and you are done.

Step 2: If $\theta$ is outside the range, replace it by the co-valued angle inside.

Find the unique $\phi \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ with $\sin\phi = \sin\theta$. The reduction identities do this quickly: …

Common Mistakes

Mistake 1: Cancelling blindly to get $\sin^{-1}(\sin\theta) = \theta$.

Why it's wrong: this only holds when $\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. Here $\frac{2\pi}{3}$ is outside that range, so writing $\frac{2\pi}{3}$ is incorrect. Correct approach: replace $\theta$ with the co-valued angle inside the range using $\sin^{-1}(\sin\theta) = \pi - \theta$, giving $\frac{\pi}{3}$. …