Q.Find the principal value of $\sin^{-1}\left(\dfrac{1}{\sqrt{2}}\right)$.

The key idea is that the principal value of $\sin^{-1}x$ lies in the closed interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

We need the angle $\theta$ in this range such that $\sin\theta = \frac{1}{\sqrt{2}}$.

Since $\sin\frac{\pi}{4} = \frac{1}{\sqrt{2}}$ and $\frac{\pi}{4}$ lies within $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, this is the required principal value.

The key to solving this lies in understanding what "principal value" means for inverse trigonometric functions. Unlike the regular sine function, which is periodic and gives the same output for infinitely many inputs, the inverse sine (or arcsine) is defined to give a single, unique output for each input. This is done by restricting the domain of the sine function so that it becomes one-to-one.

For $\sin^{-1} x$, the output (the angle) is always chosen from the interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. This is the principal value branch. So when we ask for $\sin^{-1}\left(\frac{1}{\sqrt{2}}\right)$, we are really asking: "Which angle in the range $[-\frac{\pi}{2}, \frac{\pi}{2}]$ has a sine equal to $\frac{1}{\sqrt{2}}$?"

Let’s work through it step by step.

1. Recall the standard sine values.

   You know that $\sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$. This is a standard angle from the unit circle. So $\frac{\pi}{4}$ is a candidate.

2. Check if $\frac{\pi}{4}$ lies in the principal value range.

   The principal value range for $\sin^{-1}$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$. Since $\frac{\pi}{4}$ is positive and less than $\frac{\pi}{2}$, it falls comfortably inside this interval.

3. Are there other angles with the same sine?

   Yes — for example, $\sin\left(\frac{3\pi}{4}\right) = \frac{1}{\sqrt{2}}$ as well. But $\frac{3\pi}{4}$ is outside the principal range (it’s greater than $\frac{\pi}{2}$), so it is not the principal value. Similarly, angles like $\frac{9\pi}{4}$ or $-\frac{7\pi}{4}$ also work, but none of them lie in $[-\frac{\pi}{2}, \frac{\pi}{2}]$ except $\frac{\pi}{4}$.

4. Confirm the value. $\frac{1}{\sqrt{2}}$ is positive, so the principal angle must be in the first quadrant (where sine is positive) of the restricted domain. The only such angle in $[0, \frac{\pi}{2}]$ with that sine is $\frac{\pi}{4}$.

Method: Reading a principal value from the standard-angle table

Use this whenever a question asks for the principal value of a single inverse-trig expression like $\sin^{-1}(k)$, $\cos^{-1}(k)$ or $\tan^{-1}(k)$ where $k$ is a standard value.

Steps

Step 1: Write down the principal-value range for this particular inverse function.

The answer is forced to live in one fixed interval. For inverse sine it is

Writing the range first is what stops you from picking a wrong co-terminal angle later.

Step 2: Rephrase the question as "which angle in that range gives this value?"

Instead of "what is $\sin^{-1}(k)$?", ask "for which $\theta$ in the range is $\sin\theta = k$?" Pull $\theta$ from the standard angles ($0,\tfrac{\pi}{6},\tfrac{\pi}{4},\tfrac{\pi}{3},\tfrac{\pi}{2}$) and their negatives.

Step 3: Confirm the candidate angle actually lies inside the range.

Several angles share the same sine; only one sits in $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$. Accept only that one. A positive argument gives an answer in $\left(0,\tfrac{\pi}{2}\right]$; a negative argument gives one in $\left[-\tfrac{\pi}{2},0\right)$. That quick sign check catches most slips.

Common Mistakes

Mistake 1: Giving $\dfrac{3\pi}{4}$ instead of $\dfrac{\pi}{4}$.

Why it's wrong: $\sin\tfrac{3\pi}{4}=\tfrac{1}{\sqrt2}$ too, so it looks valid, but $\tfrac{3\pi}{4}>\tfrac{\pi}{2}$ lies outside the principal range $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$. Correct approach: among all angles with sine $\tfrac{1}{\sqrt2}$, accept only the one inside the range, which is $\tfrac{\pi}{4}$.

Mistake 2: Reading $\sin^{-1}$ as a reciprocal.

Why it's wrong: the $-1$ means "inverse function", not a power, so $\sin^{-1}x \ne \tfrac{1}{\sin x}=\csc x$. Correct approach: treat $\sin^{-1}(1/\sqrt2)$ as "the angle whose sine is $1/\sqrt2$", giving an angle $\tfrac{\pi}{4}$, not a ratio.