Q.Find the principal value of the following: $\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)$

The key idea is that the principal value of $\cos^{-1}$ lies in $[0, \pi]$, and we must pick the angle in that range whose cosine equals $\frac{\sqrt{3}}{2}$.

We know $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$. Since $\frac{\pi}{6}$ lies in $[0, \pi]$, it is the principal value.

No other angle in $[0, \pi]$ gives the same cosine, so this is unique.

The key to solving any inverse trigonometric problem is remembering the range of the principal value branch. For $\cos^{-1}$, the output is always an angle in the closed interval $[0, \pi]$. This is not arbitrary — it’s the standard choice that makes the function one-to-one and therefore invertible.

Why $[0, \pi]$? Because cosine is strictly decreasing from $1$ to $-1$ as the angle goes from $0$ to $\pi$. So every possible cosine value (between $-1$ and $1$) corresponds to exactly one angle in that interval. That’s what we mean by “principal value.”

Now, the problem asks for $\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)$. We need the angle $\theta$ in $[0, \pi]$ such that $\cos \theta = \frac{\sqrt{3}}{2}$.

1. Recall the standard cosine values.

   From the unit circle or the $30^\circ$-$60^\circ$-$90^\circ$ triangle, we know:

   $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$

   $\cos\left(\frac{11\pi}{6}\right) = \frac{\sqrt{3}}{2}$ as well, but $\frac{11\pi}{6}$ is not in $[0, \pi]$.

2. Check the range.

   $\frac{\pi}{6}$ is about $30^\circ$, which lies comfortably inside $[0, \pi]$. So it’s a valid principal value.

3. Eliminate other candidates.

   Could $\theta = -\frac{\pi}{6}$ work? $\cos(-\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$, but $-\frac{\pi}{6}$ is not in $[0, \pi]$. So it’s rejected.

   Could $\theta = \frac{11\pi}{6}$? Same cosine, but $330^\circ$ is outside $[0, \pi]$. Rejected.

4. Confirm the value. $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$ exactly, and $\frac{\pi}{6} \in [0, \pi]$. So the principal value is $\frac{\pi}{6}$.

Method: Evaluating $\cos^{-1}$ of a standard positive value

Steps

Step 1: Recall the range $[0,\pi]$.

Step 2: Match to a standard angle whose cosine is the given value: $\cos\tfrac{\pi}{6}=\tfrac{\sqrt3}{2}$.

Step 3: Confirm uniqueness.

Check the angle lies in $[0,\pi]$ (here $\tfrac{\pi}{6}$ does). Cosine is one-to-one on $[0,\pi]$, so exactly one angle in the range shares this cosine.

Common Mistakes

Mistake 1: Giving $-\frac{\pi}{6}$ or $\frac{11\pi}{6}$.

Why it's wrong: both have cosine $\frac{\sqrt3}{2}$ but lie outside $[0,\pi]$. Correct approach: keep the unique in-range angle $\frac{\pi}{6}$.

Mistake 2: Confusing $\cos^{-1}$ with $\sin^{-1}$.

Why it's wrong: $\sin^{-1}\frac{\sqrt3}{2}=\frac{\pi}{3}$, a different angle. Correct approach: match the value to the correct ratio in the correct range.