Animated Solution for Mathematics - Inverse Trigonometric Functions: The trigonometric equation ${\sin }^{-1}x=2{\sin }^{-1}a$ has a solution for

Imagine you are standing before a gate. On one side, you have the expression ${\sin }^{-1}x$, and on the other, $2{\sin }^{-1}a$. They are locked in an equality: ${\sin }^{-1}x=2{\sin }^{-1}a$.

This equality is not a universal truth. It is a fragile condition that only holds when the values of $a$ allow it to exist.

To solve this, we must first respect the 'safe zone' of the inverse sine function. The function $y={\sin }^{-1}x$ is a master of restraint. Its output is strictly bound to the principal value branch, which is the closed interval $[-\frac{\pi }{2},\frac{\pi }{2}]$.

Since the left-hand side of our equation, ${\sin }^{-1}x$, is trapped within $[-\frac{\pi }{2},\frac{\pi }{2}]$, the right-hand side, $2{\sin }^{-1}a$, must also be trapped within those exact same walls. This is the bridge that connects the two sides.

We can write this as a formal constraint:

This is not just an algebraic step; it is a physical necessity. If $2{\sin }^{-1}a$ were to exceed $\frac{\pi }{2}$, the equation would have no solution because ${\sin }^{-1}x$ could never reach that value.

Now, let us perform the surgery. We want to isolate ${\sin }^{-1}a$. We divide the entire inequality by $2$, which gives us:

Notice what has happened here. Our acceptable region on the y-axis has shrunk significantly. We are no longer looking at the full range of the inverse sine function; we are looking at a narrow slice.

To find the range of $a$, we must now 'undo' the inverse sine. We apply the sine function to all parts of the inequality:

Because the sine function is strictly increasing in this interval, the inequality signs remain perfectly intact.

We know that $\sin (\frac{\pi }{4})=\frac{1}{\sqrt{2}}$ and, because sine is an odd function, $\sin (-\frac{\pi }{4})=-\frac{1}{\sqrt{2}}$. Substituting these values back, we arrive at the final, elegant condition:

In the language of absolute values, this is simply $|a|\le \frac{1}{\sqrt{2}}$.

This is the exact range of $a$ that keeps our equation valid. It is a beautiful result, showing how a seemingly complex equation is governed by the simple, rigid boundaries of its own functions.