Animated Solution for Mathematics - Inverse Trigonometric Functions: For $\alpha ,\beta ,\gamma \ne 0$. If ${\sin }^{-1}\alpha +{\sin }^{-1}\beta +{\sin }^{-1}\gamma =\pi$ and $(\alpha +\beta +\gamma )(\alpha -\gamma +\beta )=3\alpha \beta$, then $\gamma$ equal to

Welcome, future engineer. Today, we are going to peel back the layers of a problem that looks like a dense algebraic nightmare but is actually a beautiful, elegant geometric puzzle.

When you first see ${\sin }^{-1}\alpha +{\sin }^{-1}\beta +{\sin }^{-1}\gamma =\pi$, your instinct might be to reach for complex trigonometric identities. Resist that urge. Instead, pause and look at the structure.

What do we know about the sum of three angles equaling $\pi$? It is the hallmark of a triangle. Let us define $A={\sin }^{-1}\alpha$, $B={\sin }^{-1}\beta$, and $C={\sin }^{-1}\gamma$.

Suddenly, we are not just doing algebra; we are standing inside a triangle with angles $A,B,$ and $C$. This means $\alpha =\sin A$, $\beta =\sin B$, and $\gamma =\sin C$. We have just unlocked the door.

Now that we have established our triangle, we need to relate these sines to the sides of the triangle. The Sine Rule is our best friend here:

Since $\sin A=\alpha$, $\sin B=\beta$, and $\sin C=\gamma$, we can see that the sides $a,b,$ and $c$ are directly proportional to $\alpha ,\beta ,$ and $\gamma$.

This is a powerful realization. It means that for the purpose of ratios, we can treat $\alpha ,\beta ,$ and $\gamma$ as the side lengths of our triangle. The geometry is now working for us, not against us.

Look at the second equation: $(\alpha +\beta +\gamma )(\alpha -\gamma +\beta )=3\alpha \beta$. It looks messy, but look closer.

If we group $(\alpha +\beta )$ together, we see the structure $(x+y)(x-y)$, where $x=\alpha +\beta$ and $y=\gamma$. This is the classic difference of squares identity!

Expanding this, we get:

Now, expand the perfect square: ${\alpha }^2+{\beta }^2+2\alpha \beta -{\gamma }^2=3\alpha \beta$. Subtracting $2\alpha \beta$ from both sides gives us the clean, beautiful result:

We have tamed the beast.

We are almost there. We have ${\alpha }^2+{\beta }^2-{\gamma }^2=\alpha \beta$. Does this look like anything you have seen before?

It is almost the Cosine Rule! The Cosine Rule states:

If we divide our equation by $2\alpha \beta$, we get:

This is exactly $\cos C=\frac{1}{2}$. Since $C$ is an angle in a triangle, $C=\frac{\pi }{3}$ or $60^{\circ }$.

Finally, we return to our definition: $\gamma =\sin C=\sin (\frac{\pi }{3})=\frac{\sqrt{3}}{2}$.

You have successfully navigated the intersection of algebra and geometry. The final answer is $\gamma =\frac{\sqrt{3}}{2}$. Take a moment to appreciate how the complexity collapsed into such a simple, elegant value.