Q.The value of $\cos^{-1}\left(\cos\frac{3\pi}{2}\right)$ is equal to
(A) $\frac{\pi}{2}$
(B) $\frac{3\pi}{2}$
(C) $\frac{5\pi}{2}$
(D) $\frac{7\pi}{2}$

Concept: Principal value branch of $\cos^{-1}$ — the output must lie in $[0, \pi]$.

Step 1: First evaluate the inner cosine.

$\cos\frac{3\pi}{2} = 0$.

Step 2: So the problem reduces to $\cos^{-1}(0)$. …

The problem looks simple, but it traps students who forget the definition of the inverse cosine function. $\cos^{-1}$ is not the "undo" button for every angle — it only gives an output in a specific range, called the principal value branch.

For $\cos^{-1}$, that range is $[0, \pi]$. So no matter what angle you feed into $\cos$, when you take $\cos^{-1}$ of the result, you must get back an angle that lies between $0$ and $\pi$ inclusive.

Now let's work through it.

1. First, evaluate the inner expression: $\cos\frac{3\pi}{2}$.

   $\frac{3\pi}{2}$ is $270^\circ$, which lies on the negative y-axis. The cosine of $270^\circ$ is $0$.

   So $\cos\frac{3\pi}{2} = 0$.

2. The problem reduces to finding $\cos^{-1}(0)$.

   We need the angle $\theta$ in $[0, \pi]$ such that $\cos\theta = 0$.

3. On the unit circle, cosine is zero at $\frac{\pi}{2}$ and $\frac{3\pi}{2}$. But only $\frac{\pi}{2}$ lies in $[0, \pi]$.

   Therefore $\cos^{-1}(0) = \frac{\pi}{2}$. …

Method: Evaluating $\cos^{-1}(\cos x)$ when $x$ is outside $[0,\pi]$

You cannot cancel $\cos^{-1}$ and $\cos$ directly unless the angle already sits in the principal range. The method is to first collapse to a plain cosine value, then invert into $[0,\pi]$.

Steps

Step 1: Check whether $x$ is already in the principal range.

$\cos^{-1}(\cos x)=x$ holds only when $x\in[0,\pi]$. If $x$ lies outside this interval (as with $\tfrac{3\pi}{2}$), the shortcut is invalid — do not cancel.

Step 2: Evaluate the inner cosine to a number.

Compute $\cos x$ as an ordinary value using the unit circle (e.g. $\cos\tfrac{3\pi}{2}=0$). This turns the problem into $\cos^{-1}(\text{number})$.

Step 3: Invert into $[0,\pi]$. …

Common Mistakes

Mistake 1: Cancelling to get $\cos^{-1}\left(\cos\tfrac{3\pi}{2}\right)=\tfrac{3\pi}{2}$.

Why it's wrong: $\cos^{-1}(\cos x)=x$ only when $x\in[0,\pi]$, and $\tfrac{3\pi}{2}$ is outside it — indeed $\tfrac{3\pi}{2}$ isn't even a valid $\cos^{-1}$ output. Correct approach: evaluate $\cos\tfrac{3\pi}{2}=0$ first, then $\cos^{-1}(0)=\tfrac{\pi}{2}$. …