Animated Solution for Mathematics - Inverse Trigonometric Functions: Considering the principal values of the inverse trigonometric functions, ${\sin }^{-1}\left(\frac{\sqrt{3}}{2}x+\frac{1}{2}\sqrt{1-x^2}\right),-\frac{1}{2}<x<\frac{1}{\sqrt{2}}$, is equal to

Unlocking the Mystery

Welcome, future engineer. Today, we are going to demystify a problem that often intimidates students at first glance.

We are looking at the expression:

This is defined within the domain $-\frac{1}{2}<x<\frac{1}{\sqrt{2}}$. At first, this might look like a chaotic mess of radicals and inverse functions, but I want you to see it as a puzzle where every piece has a specific place.

The Substitution Strategy

Whenever you see $\sqrt{1-x^2}$ in a trigonometric context, your mathematical intuition should immediately scream, "Substitution!" This term is a dead giveaway for the identity ${\sin }^2\theta +{\cos }^2\theta =1$.

By setting $x=\sin \theta$, we transform the algebraic expression into a trigonometric one. This implies $\theta ={\sin }^{-1}x$.

Now, let us look at the domain. If $x$ is between $-\frac{1}{2}$ and $\frac{1}{\sqrt{2}}$, then $\sin \theta$ must also be in that same interval. Looking at our unit circle, this restricts $\theta$ to the interval $(-\frac{\pi }{6},\frac{\pi }{4})$.

The Identity Reveal

With our substitution $x=\sin \theta$, the term $\sqrt{1-x^2}$ becomes $\sqrt{1-{\sin }^2\theta }$, which simplifies to $\sqrt{{\cos }^2\theta }$, or $|\cos \theta |$.

Because our $\theta$ is restricted to $(-\frac{\pi }{6},\frac{\pi }{4})$, we are firmly in the first and fourth quadrants, where cosine is always positive. Thus, we can confidently say $\sqrt{1-x^2}=\cos \theta$.

Now, substitute these back into our original expression:

Do you recognize those coefficients? $\frac{\sqrt{3}}{2}$ and $\frac{1}{2}$ are $\cos \frac{\pi }{6}$ and $\sin \frac{\pi }{6}$, respectively.

The expression now reads:

This is the classic sine addition formula, $\sin (A+B)=\sin A\cos B+\cos A\sin B$, in disguise! The entire bracket collapses beautifully into $\sin (\theta +\frac{\pi }{6})$.

The Final Gatekeeper

We are left with ${\sin }^{-1}(\sin (\theta +\frac{\pi }{6}))$. Many students rush here and simply write $\theta +\frac{\pi }{6}$.

But wait! As your mentor, I must remind you: never skip the principal range check. We know $\theta \in (-\frac{\pi }{6},\frac{\pi }{4})$.

If we add $\frac{\pi }{6}$ to every part of this inequality, we get:

This simplifies to $0<\theta +\frac{\pi }{6}<\frac{5\pi }{12}$. Since $\frac{5\pi }{12}$ is approximately $75^{\circ }$, which is well within the principal range of $[-\frac{\pi }{2},\frac{\pi }{2}]$, we are safe to cancel the inverse sine and the sine.

The expression simplifies to $\theta +\frac{\pi }{6}$.

Finally, we substitute back our initial assumption, $\theta ={\sin }^{-1}x$.

Our result is:

See how the complexity vanished? By trusting the substitution, respecting the domain, and verifying the principal range, we turned a terrifying expression into a simple, elegant answer.