Q.$\cos^{-1}\left(\cos\frac{7\pi}{6}\right)$ is equal to (A) $\frac{7\pi}{6}$ (B) $\frac{5\pi}{6}$ (C) $\frac{\pi}{3}$ (D) $\frac{\pi}{6}$

Concept: Inverse Trigonometric Graphs — the principal value branch of $\cos^{-1}$ is $[0, \pi]$, so the output must lie in this interval.

Step 1: Compute the inner value.

$\cos\frac{7\pi}{6} = \cos\left(\pi + \frac{\pi}{6}\right) = -\cos\frac{\pi}{6} = -\frac{\sqrt{3}}{2}$.

Step 2: We need $\cos^{-1}\left(-\frac{\sqrt{3}}{2}\right)$.

Let $\theta = \cos^{-1}\left(-\frac{\sqrt{3}}{2}\right)$. Then $\cos\theta = -\frac{\sqrt{3}}{2}$ and $\theta \in [0, \pi]$. …

1. Understand the function $\cos^{-1}(\cos x)$

   The inverse cosine function, $\cos^{-1}$, is defined to give an output only in the principal value branch $[0,\pi]$. For any real $x$, $\cos^{-1}(\cos x)$ equals the unique angle $\theta \in [0,\pi]$ such that $\cos\theta = \cos x$.

   So the problem reduces to: Find the angle in $[0,\pi]$ whose cosine is the same as $\cos\frac{7\pi}{6}$.

2. Evaluate $\cos\frac{7\pi}{6}$

   $\frac{7\pi}{6} = \pi + \frac{\pi}{6}$. Using the cosine addition formula:

So we need an angle $\theta \in [0,\pi]$ such that $\cos\theta = -\frac{\sqrt{3}}{2}$.

3. Find $\theta$ in $[0,\pi]$ with that cosine …

Method: Evaluating $\cos^{-1}(\cos\theta)$

Use this to simplify $\cos^{-1}(\cos\theta)$ when $\theta$ may be outside the principal range of $\cos^{-1}$.

Steps

Step 1: Recall the principal range.

$\cos^{-1}$ returns an angle in $[0, \pi]$. If $\theta$ already lies there, the answer is $\theta$.

Step 2: If $\theta$ is outside, find the co-valued angle in $[0,\pi]$.

Because cosine is even and $2\pi$-periodic, use whichever reduction applies:

or first reduce $\theta$ modulo $2\pi$ if it exceeds $2\pi$. …

Common Mistakes

Mistake 1: Cancelling to get $\cos^{-1}\left(\cos\frac{7\pi}{6}\right) = \frac{7\pi}{6}$.

Why it's wrong: $\frac{7\pi}{6} \approx 210^\circ$ is outside the principal range $[0,\pi]$, so it cannot be the output. Correct approach: find the angle in $[0,\pi]$ with the same cosine, namely $\frac{5\pi}{6}$.

Mistake 2: Mishandling the sign of $\cos\frac{7\pi}{6}$. …