Q.The value of $\sin^{-1}\left(\sin\frac{3\pi}{5}\right)$ is __________.

The key idea is that $\sin^{-1}(\sin x)$ does not always return $x$ — it returns the principal value, which lies in $[-\frac{\pi}{2}, \frac{\pi}{2}]$. So we must first adjust the given angle to fall within that range.

Step 1: Note that $\frac{3\pi}{5} = 108^\circ$, which is outside $[-\frac{\pi}{2}, \frac{\pi}{2}]$. …

The key here is understanding what $\sin^{-1}$ actually does. It’s not a simple cancellation — it asks: “Give me the angle in the principal range $[-\pi/2, \pi/2]$ whose sine equals the given number.” So when you see $\sin^{-1}(\sin x)$, the result is not always $x$. It’s $x$ only if $x$ already lies in $[-\pi/2, \pi/2]$.

Here, $x = \frac{3\pi}{5}$. Let’s check: $\frac{3\pi}{5} = 108^\circ$, which is clearly outside $[-90^\circ, 90^\circ]$. So we need to find another angle — call it $\theta$ — such that:

• $\sin \theta = \sin \frac{3\pi}{5}$, and
• $\theta \in [-\pi/2, \pi/2]$.

Let’s work through it.

1. Find the reference angle.

   $\frac{3\pi}{5}$ is in the second quadrant (since $\pi/2 < 3\pi/5 < \pi$). In the second quadrant, sine is positive. The reference angle is $\pi - \frac{3\pi}{5} = \frac{2\pi}{5}$.

2. Use the sine symmetry.

   For any angle in the second quadrant, $\sin(\pi - \alpha) = \sin \alpha$. So:

3. Check the principal range.

   $\frac{2\pi}{5} = 72^\circ$. Is this in $[-\pi/2, \pi/2]$? Yes — $72^\circ$ is less than $90^\circ$ and greater than $-90^\circ$. So $\frac{2\pi}{5}$ is a valid principal value.

4. No other candidate works. …

Method: Simplifying $\sin^{-1}(\sin x)$

Apply this whenever the inner angle $x$ may lie outside the principal range $[-\tfrac{\pi}{2},\tfrac{\pi}{2}]$.

Steps

Step 1: Check whether $x$ is already in $[-\tfrac{\pi}{2},\tfrac{\pi}{2}]$.

If it is, then $\sin^{-1}(\sin x)=x$ and you are finished.

Step 2: If not, find an angle with the same sine that IS in range. …

Common Mistakes

Mistake 1: Cancelling to get $\frac{3\pi}{5}$.

Why it's wrong: $\sin^{-1}(\sin x)=x$ only when $x\in[-\frac{\pi}{2},\frac{\pi}{2}]$, and $\frac{3\pi}{5}$ (108°) is outside it. Correct approach: replace it by the in-range angle $\pi-\frac{3\pi}{5}=\frac{2\pi}{5}$.

Mistake 2: Using $x-\pi$ or $\pi+x$ instead of $\pi-x$. …