Animated Solution for Mathematics - Inverse Trigonometric Functions: The principal value of ${\sin }^{-1}\left(\sin \frac{2\pi }{3}\right)$ is

Imagine you are standing at the threshold of a complex trigonometric problem. You see ${\sin }^{-1}\left(\sin \frac{2\pi }{3}\right)$.

Your brain, wired for efficiency, screams: "Cancel them! Just cross out the sine and the inverse sine and walk away with $\frac{2\pi }{3}$!" It feels like the most logical, elegant step.

But in the world of JEE Advanced, this is the siren song that leads many brilliant students into a trap. Today, we are going to dismantle that trap and understand the beautiful, rigid geometry that governs inverse trigonometry.

The inverse sine function, ${\sin }^{-1}(x)$, is not just a simple "undo" button for the sine function. It is a function with a very specific, non-negotiable job.

To be a valid function, it must be "one-to-one," meaning every input must map to exactly one output. However, the sine wave is periodic—it repeats itself infinitely.

If we didn't restrict the output, ${\sin }^{-1}(x)$ would have infinitely many possible answers for a single input. To solve this, mathematicians defined the "Principal Value Branch."

Think of this as a "Red Zone" or a "Safe Zone" on the graph. For ${\sin }^{-1}(x)$, this zone is strictly defined as:

Any output from the inverse sine function MUST land in this interval. If your calculation lands outside this, you haven't found the principal value; you've found an imposter.

Let's look at our angle, $\frac{2\pi }{3}$. In degrees, this is $120^{\circ }$.

If you visualize the unit circle, $120^{\circ }$ is in the second quadrant. Our "Safe Zone" is the first and fourth quadrants (from $-90^{\circ }$ to $90^{\circ }$).

Clearly, $120^{\circ }$ is way out in the wilderness, far beyond the $90^{\circ }$ boundary. This is why the blind cancellation fails.

The function ${\sin }^{-1}$ is looking for an angle in the safe zone that produces the same sine value as $\sin (\frac{2\pi }{3})$. It doesn't care about the angle $120^{\circ }$ itself; it cares about the value of the sine at that point.

Now, how do we bring this rogue angle back into the fold? We need an angle $\alpha$ such that $\sin (\alpha )=\sin (\frac{2\pi }{3})$ and $\alpha \in [-\frac{\pi }{2},\frac{\pi }{2}]$.

This is where the beauty of allied angles shines. We know the identity:

This identity is a bridge. It tells us that the sine of an angle in the second quadrant is identical to the sine of its supplement in the first quadrant.

So, we rewrite $\frac{2\pi }{3}$ as $\pi -\frac{\pi }{3}$. Now, our expression becomes:

By our identity, this is equivalent to ${\sin }^{-1}\left(\sin \frac{\pi }{3}\right)$.

Look at $\frac{\pi }{3}$ (or $60^{\circ }$). It is comfortably nestled within our principal branch $[-\frac{\pi }{2},\frac{\pi }{2}]$.

Now, and ONLY now, can we perform the cancellation. The inverse sine and the sine neutralize each other, leaving us with the elegant, correct answer:

You see, the math didn't change; we just navigated the geometry correctly. The next time you face an inverse trigonometric expression, don't rush to cancel. Pause, check the "Safe Zone," and use your identities to guide the angle home. You've got this!