Q.Find the value of $\sin^{-1}\left(\sin\dfrac{3\pi}{5}\right)$.

The key idea is that $\sin^{-1}(\sin x)$ does not always return $x$ — it returns the principal value of the inverse sine, which lies in $[-\frac{\pi}{2}, \frac{\pi}{2}]$. So we must first adjust the angle $\frac{3\pi}{5}$ to an equivalent angle within that range.

1. $\frac{3\pi}{5} = 108^\circ$, which is in the second quadrant. Sine is positive there.
2. Using $\sin(\pi - \theta) = \sin\theta$, we have $\sin\frac{3\pi}{5} = \sin\left(\pi - \frac{3\pi}{5}\right) = \sin\frac{2\pi}{5}$. …

1. The core concept: what $\sin^{-1}(\sin x)$ actually does

The function $\sin^{-1}$ (also written $\arcsin$) is defined as the inverse of $\sin$, but only on the restricted domain $[-\pi/2, \pi/2]$. This means:

If $x$ lies outside this interval, $\sin^{-1}(\sin x)$ returns the unique angle $y$ in $[-\pi/2, \pi/2]$ such that $\sin y = \sin x$. That $y$ is called the principal value.

So the problem reduces to: Find an angle in $[-\pi/2, \pi/2]$ whose sine equals $\sin(3\pi/5)$.

2. Where does $3\pi/5$ sit on the unit circle?

$3\pi/5$ radians is $108^\circ$. That’s in Quadrant II (between $\pi/2$ and $\pi$). In this quadrant, sine is positive, but the angle itself is far outside $[-\pi/2, \pi/2]$.

We need a reference angle approach: any two angles with the same sine are either:

• symmetric about the y-axis (i.e., $\theta$ and $\pi - \theta$), or
• differ by multiples of $2\pi$.

Since $3\pi/5$ is in Quadrant II, its reference angle (the acute angle it makes with the x-axis) is:

And $\sin(3\pi/5) = \sin(2\pi/5)$. …

Method: Evaluating $\sin^{-1}(\sin x)$ when $x$ is outside the principal range

Use this for any $f^{-1}(f(x))$ evaluation ($\sin^{-1}(\sin x)$, $\cos^{-1}(\cos x)$, etc.) where the inner angle does not already lie in the inverse function's range.

Steps

Step 1: Check whether the inner angle is already in the principal range.

For inverse sine the range is $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$. If $x$ is inside it, the answer is simply $x$ and you are done. If not — as when $x=\tfrac{3\pi}{5}>\tfrac{\pi}{2}$ — do not stop at $x$.

Step 2: Replace $x$ by an equivalent angle with the SAME sine that lies in the range.

Use the identity that preserves the value:

$\sin(\pi - \theta) = \sin\theta.$ …

Common Mistakes

Mistake 1: Answering $\dfrac{3\pi}{5}$ directly.

Why it's wrong: $\sin^{-1}(\sin x)=x$ only when $x\in\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$, and $\tfrac{3\pi}{5}>\tfrac{\pi}{2}$ is outside it. Correct approach: replace $\tfrac{3\pi}{5}$ with an angle of equal sine that lies in the range.

Mistake 2: Using $2\pi-\dfrac{3\pi}{5}$ or $\dfrac{3\pi}{5}-\pi$ instead of $\pi-\dfrac{3\pi}{5}$. …