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

The key idea is that the principal value of $\sin^{-1}x$ lies in the closed interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, and we need the angle whose sine is $-\frac12$.

1. We know $\sin\left(\frac{\pi}{6}\right) = \frac12$, so for a negative sine, the angle must be negative in this range.
2. Since $\sin(-\theta) = -\sin\theta$, we have $\sin\left(-\frac{\pi}{6}\right) = -\frac12$.
3. The angle $-\frac{\pi}{6}$ lies within $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, so it is the principal value.

The key to solving this lies in understanding what "principal value" means for inverse trigonometric functions. Unlike regular sine, which is many-to-one (infinitely many angles give the same sine), the inverse sine $\sin^{-1}(x)$ is defined as a function — it must give exactly one output for each input. To make this work, we restrict the range of $\sin^{-1}$ to a specific interval where sine is one-to-one.

For $\sin^{-1}$, that interval is $[-\frac{\pi}{2}, \frac{\pi}{2}]$. So when we ask for the principal value of $\sin^{-1}\left(-\frac{1}{2}\right)$, we are looking for the unique angle $\theta$ in that interval whose sine is $-\frac{1}{2}$.

Let’s find the correct angle step by step.

1. Recall the standard sine values.

   We know $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$. Since sine is an odd function ($\sin(-\theta) = -\sin\theta$), we have $\sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2}$.

2. Check the range.

   The angle $-\frac{\pi}{6}$ is approximately $-0.5236$ radians. This lies squarely within $[-\frac{\pi}{2}, \frac{\pi}{2}]$ because $-\frac{\pi}{2} \approx -1.5708$ and $\frac{\pi}{2} \approx 1.5708$. So it qualifies as a candidate.

3. Confirm uniqueness.

   Could there be another angle in $[-\frac{\pi}{2}, \frac{\pi}{2}]$ with the same sine? No — within this interval, sine is strictly increasing (from $-1$ at $-\frac{\pi}{2}$ to $1$ at $\frac{\pi}{2}$), so each sine value corresponds to exactly one angle. Therefore $-\frac{\pi}{6}$ is the only possibility.

Thus, the principal value is $-\frac{\pi}{6}$.

Method: Finding a principal value of $\sin^{-1}$

This is the standard technique for any "principal value of $\sin^{-1}(k)$" question — you locate the one angle in the fixed range whose sine is the given number.

Steps

Step 1: Write down the principal range.

By definition the answer must lie in

Step 2: Find the reference angle from the size of the number.

Ignore the sign for a moment and recall the standard value: for $|x|=\tfrac12$ the reference angle is $\tfrac{\pi}{6}$, for $\tfrac{1}{\sqrt2}$ it is $\tfrac{\pi}{4}$, for $\tfrac{\sqrt3}{2}$ it is $\tfrac{\pi}{3}$.

Step 3: Attach the correct sign.

Because $\sin$ is odd, a negative argument gives a negative angle: $\sin^{-1}(-a)=-\sin^{-1}(a)$. A positive argument gives a positive angle. The answer is never obtained by adding $\pi$ or $2\pi$.

Step 4: Check it lies in $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$.

If your candidate angle sits outside this interval, it is not the principal value — replace it with the co-terminal/equivalent angle that does.

Common Mistakes

Mistake 1: Giving $+\tfrac{\pi}{6}$ instead of $-\tfrac{\pi}{6}$.

Why it's wrong: the argument $-\tfrac12$ is negative, and $\sin$ is odd, so the principal value must be negative. Correct approach: use $\sin^{-1}(-a)=-\sin^{-1}(a)$, giving $-\tfrac{\pi}{6}$.

Mistake 2: Choosing an angle outside $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$, such as $\tfrac{7\pi}{6}$ or $\tfrac{11\pi}{6}$.

Why it's wrong: those angles do have sine $-\tfrac12$, but they are not principal values. Correct approach: only the angle inside the closed range $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$ counts, which is $-\tfrac{\pi}{6}$.

Mistake 3: Reading $\sin^{-1}$ as $\tfrac{1}{\sin}$.

Why it's wrong: the "$-1$" denotes the inverse function, not a reciprocal power. Correct approach: treat $\sin^{-1}(-\tfrac12)$ as "the angle whose sine is $-\tfrac12$", not as $\csc(-\tfrac12)$.