Q.If $\sin^{-1} x = y$, then (A) $0 \le y \le \pi$ (B) $-\frac{\pi}{2} \le y \le \frac{\pi}{2}$ (C) $0 < y < \pi$ (D) $-\frac{\pi}{2} < y < \frac{\pi}{2}$

Concept: Inverse Sine Principal Value

The inverse sine function, $\sin^{-1} x$, is defined to return the principal value — the unique angle in the restricted range where sine is one-to-one.

For $\sin^{-1} x = y$, the standard convention is:

• $y$ must lie in the closed interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
• This includes both endpoints because $\sin(-\frac{\pi}{2}) = -1$ and $\sin(\frac{\pi}{2}) = 1$.

Step 1: The sine function is one-to-one only on $[-\frac{\pi}{2}, \frac{\pi}{2}]$. …

The question tests a definition that every student memorises but often misremembers. The inverse sine function, $\sin^{-1} x$ (also written as $\arcsin x$), is not the same as "the angle whose sine is $x$" — because that would give infinitely many angles. Instead, we restrict the range to a single, convenient interval so that the function becomes one-to-one and well-defined. That restricted range is called the principal value branch.

Why $[-\frac{\pi}{2}, \frac{\pi}{2}]$? Because on this interval, the sine function is strictly increasing (so it passes the horizontal line test) and covers all possible output values from $-1$ to $1$. No other interval of length $\pi$ does this as neatly — for instance, $[0, \pi]$ would include angles where sine is positive then negative, but the function would not be one-to-one there (since $\sin(\pi/6) = \sin(5\pi/6)$).

Now let's check each option:

1. Option (A): $0 \le y \le \pi$

   This is the principal value branch for $\cos^{-1} x$, not $\sin^{-1} x$. For example, $\sin^{-1}(1) = \frac{\pi}{2}$, which lies in this interval, but $\sin^{-1}(-1) = -\frac{\pi}{2}$, which does not lie in $[0, \pi]$. So (A) is wrong.

2. Option (B): $-\frac{\pi}{2} \le y \le \frac{\pi}{2}$

   This is exactly the definition. Every value of $\sin^{-1} x$ falls in this closed interval. For $x = 1$, $y = \frac{\pi}{2}$; for $x = -1$, $y = -\frac{\pi}{2}$; for $x = 0$, $y = 0$. All endpoints are included. So (B) is correct.

3. Option (C): $0 < y < \pi$

   This is an open interval, so it excludes $0$ and $\pi$. But $\sin^{-1}(0) = 0$, which is excluded here. Also, $\sin^{-1}(-1) = -\frac{\pi}{2}$ is not even in this interval. So (C) is wrong. …

Method: Identifying the range of an inverse trig function

For a "which interval is the range" question, recall the defining principal branch rather than re-reasoning from the graph each time.

Steps

Step 1: Interpret the statement

$y = \sin^{-1}x$ means $y$ is the angle whose sine is $x$, drawn from the restricted set of angles chosen to make sine one-to-one.

Step 2: State the branch precisely …

Common Mistakes

Mistake 1: Choosing $0 \le y \le \pi$ (option A)

Why it's wrong: $[0,\pi]$ is the range of $\cos^{-1}$, not $\sin^{-1}$. Correct approach: $\sin^{-1}$ is centred on $0$ with range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

Mistake 2: Choosing the open interval $-\frac{\pi}{2} < y < \frac{\pi}{2}$ (option D) …