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

The key idea is that the principal value of $\cos^{-1}$ lies in the closed interval $[0, \pi]$, and we need the angle in that range whose cosine is $-\frac{1}{2}$.

Recall that $\cos(\pi/3) = \frac{1}{2}$, so for a negative cosine we look in the second quadrant. …

The inverse cosine function, $\cos^{-1}(x)$ (also written as $\arccos x$), is defined to give a single, unambiguous output for every input $x$ in $[-1, 1]$. The key is that cosine is not one-to-one over its entire domain — many angles have the same cosine. So we restrict the range of $\cos^{-1}$ to a specific interval where cosine is one-to-one and covers all possible output values.

For $\cos^{-1}$, the standard principal value range is $[0, \pi]$. This means the answer must be an angle between $0$ and $\pi$ (inclusive). Within this interval, cosine is strictly decreasing from $1$ to $-1$, so each value in $[-1, 1]$ corresponds to exactly one angle.

Now, we need the angle $\theta$ such that $\cos \theta = -\frac{1}{2}$ and $0 \le \theta \le \pi$.

1. Recall the cosine of standard angles.

   We know $\cos \frac{\pi}{3} = \frac{1}{2}$. Cosine is negative in the second quadrant (angles between $\frac{\pi}{2}$ and $\pi$). The reference angle for $\frac{1}{2}$ is $\frac{\pi}{3}$, so the angle in the second quadrant with cosine $-\frac{1}{2}$ is $\pi - \frac{\pi}{3} = \frac{2\pi}{3}$.

2. Check the principal value range.

   $\frac{2\pi}{3}$ lies in $[0, \pi]$ — indeed, it's about $120^\circ$, which is within the allowed interval. The other angle with cosine $-\frac{1}{2}$ is $-\frac{2\pi}{3}$ (or $\frac{4\pi}{3}$), but those are outside $[0, \pi]$, so they are not principal values.

3. Verify directly. …

Method: Principal value of an inverse cosine

Apply this when evaluating $\cos^{-1}$ of a number and reporting one angle.

Steps

Step 1: State the principal branch

$\cos^{-1}$ returns an angle in $[0, \pi]$. Cosine is positive on $\left[0, \frac{\pi}{2}\right)$ and negative on $\left(\frac{\pi}{2}, \pi\right]$, so the sign of the argument tells you which half to land in.

Step 2: Find the reference angle

Recall the standard angle whose cosine has the given magnitude (e.g. $\cos\frac{\pi}{3} = \frac{1}{2}$, $\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}$, $\cos\frac{\pi}{4} = \frac{1}{\sqrt{2}}$).

Step 3: Place it in the correct half …

Common Mistakes

Mistake 1: Answering $\frac{\pi}{3}$ and ignoring the minus sign

Why it's wrong: $\cos\frac{\pi}{3} = +\frac{1}{2}$, not $-\frac{1}{2}$. Correct approach: for a negative argument the principal value is in the second quadrant, $\pi - \frac{\pi}{3} = \frac{2\pi}{3}$.

Mistake 2: Treating $\cos^{-1}$ like an odd function and writing $-\frac{\pi}{3}$ …