Q.Find the principal value of the following: $3\sin^{-1} x = \sin^{-1} (3x - 4x^3)$, $x \in \left[-\frac{1}{2}, \frac{1}{2}\right]$

Concept: Inverse Sine Principal Value — The identity $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$ is used, but the domain must be restricted so that both sides lie in the principal range of $\sin^{-1}$, i.e. $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

Step 1: Let $\theta = \sin^{-1} x$. Then $x = \sin\theta$ and $\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

Since $x \in \left[-\frac{1}{2}, \frac{1}{2}\right]$, we have $\theta \in \left[-\frac{\pi}{6}, \frac{\pi}{6}\right]$.

Step 2: The right-hand side becomes $\sin^{-1}(3\sin\theta - 4\sin^3\theta) = \sin^{-1}(\sin 3\theta)$.

For $\theta \in \left[-\frac{\pi}{6}, \frac{\pi}{6}\right]$, we have $3\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, which lies exactly in the principal range of $\sin^{-1}$.

Step 3: Therefore $\sin^{-1}(\sin 3\theta) = 3\theta = 3\sin^{-1} x$, and the identity holds as an equality for all $x$ in the given interval.

Why this identity works — the concept

The formula $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$ is a triple-angle identity from trigonometry. If we set $\theta = \sin^{-1}x$, then $\sin\theta = x$, and the right-hand side becomes $\sin(3\theta) = 3x - 4x^3$. So we have:

Taking the inverse sine of both sides gives:

Now the left side is not simply $3\sin^{-1}x$ — it is $3\sin^{-1}x$ only if $3\sin^{-1}x$ lies in the principal range of $\sin^{-1}$, which is $\left[-\frac\pi2, \frac\pi2\right]$. If $3\sin^{-1}x$ falls outside that interval, the inverse sine "wraps" the angle back into the principal branch, and the equality fails.

So the real question is: For which $x$ does $3\sin^{-1}x$ stay inside $\left[-\frac\pi2, \frac\pi2\right]$?

Step-by-step reasoning

1. Understand the range of $\sin^{-1}x$ The principal value of $\sin^{-1}x$ is defined to lie in $\left[-\frac\pi2, \frac\pi2\right]$. So for any $x \in [-1, 1]$, we have:

2. Multiply by 3 Multiplying the inequality by 3 gives:

This is a much wider interval — it extends well beyond $\pm\frac\pi2$.

3. When does $3\sin^{-1}x$ stay inside $\left[-\frac\pi2, \frac\pi2\right]$? We need:

Divide through by 3:

Since $\sin^{-1}$ is increasing, this is equivalent to:

i.e.

4. Interpretation For $x \in \left[-\frac12, \frac12\right]$, the quantity $3\sin^{-1}x$ is guaranteed to lie in $\left[-\frac\pi2, \frac\pi2\right]$. Therefore:

and the given identity holds as a principal value.

5. What happens outside this interval? If $x > \frac12$, then $\sin^{-1}x > \frac\pi6$, so $3\sin^{-1}x > \frac\pi2$. The inverse sine then returns an angle in $\left[-\frac\pi2, \frac\pi2\right]$ that is not $3\sin^{-1}x$ but its supplement (or a shifted version). The identity would then require a different expression — for example, for $x \in \left[\frac12, 1\right]$, the correct formula is $3\sin^{-1}x = \pi - \sin^{-1}(3x - 4x^3)$.

Method: Proving a multiple-angle inverse identity via substitution

Use this whenever an inverse trig identity contains a cubic such as $3x - 4x^3$ or $4x^3 - 3x$.

Steps

Step 1: Substitute to expose the multiple angle

Set $\theta = \sin^{-1}x$ (so $x = \sin\theta$). Then $3x - 4x^3 = 3\sin\theta - 4\sin^3\theta = \sin 3\theta$ by the triple-angle identity.

Step 2: Rewrite the right-hand side

The right side becomes $\sin^{-1}(\sin 3\theta)$, which equals $3\theta$ ONLY if $3\theta$ lies in the principal branch $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

Step 3: Translate the branch condition into a domain on $x$

Require $3\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, i.e. $\theta \in \left[-\frac{\pi}{6}, \frac{\pi}{6}\right]$, i.e. $x \in \left[-\frac{1}{2}, \frac{1}{2}\right]$ — exactly the given interval. On it the identity holds.

Common Mistakes

Mistake 1: Assuming $\sin^{-1}(\sin 3\theta) = 3\theta$ for every $x$

Why it's wrong: the cancellation only works while $3\theta$ stays in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$; outside it the inverse sine folds the angle back. Correct approach: restrict to $x \in \left[-\frac{1}{2}, \frac{1}{2}\right]$, which keeps $3\theta$ in the branch.

Mistake 2: Ignoring why the interval $\left[-\frac{1}{2}, \frac{1}{2}\right]$ is given

Why it's wrong: the domain is not decorative — it is the precise condition that makes the identity valid. Correct approach: derive it from $-\frac{\pi}{2} \le 3\sin^{-1}x \le \frac{\pi}{2}$.