Q.Find the principal value of the following: $\cos^{-1} \left(\cos \frac{13\pi}{6}\right)$

Concept: Inverse Trigonometric Graphs — The principal value branch of $\cos^{-1}$ is $[0, \pi]$. We must first reduce the given angle to lie within this range.

Step 1: Simplify the inner angle.

$\frac{13\pi}{6} = 2\pi + \frac{\pi}{6}$. Since $\cos$ is periodic with period $2\pi$,

$\cos\left(\frac{13\pi}{6}\right) = \cos\left(\frac{\pi}{6}\right)$.

Step 2: Now evaluate the inverse:

$\cos^{-1}\left(\cos\frac{\pi}{6}\right)$.

Since $\frac{\pi}{6}$ lies in the principal branch $[0, \pi]$, the result is simply $\frac{\pi}{6}$.

The key to solving $\cos^{-1}(\cos \theta)$ is understanding that the inverse cosine function, $\cos^{-1}$, is not the simple inverse of $\cos$ over all real numbers. Cosine is periodic and many-to-one, so to define an inverse we restrict its domain to $[0, \pi]$. This restricted cosine is one-to-one, and its inverse, $\cos^{-1}$, gives back an angle only in $[0, \pi]$.

So when you see $\cos^{-1}(\cos x)$, the result is not automatically $x$. It is the unique angle in $[0, \pi]$ whose cosine equals $\cos x$. That angle is often called the principal value.

Let’s walk through the problem.

1. Simplify the inner cosine first. The angle $\frac{13\pi}{6}$ is large — more than $2\pi$.

Since cosine has period $2\pi$,

2. Now the problem becomes:

We need the angle $\theta$ in $[0, \pi]$ such that $\cos\theta = \frac{\sqrt{3}}{2}$.

3. Recall the standard angles. $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$, and $\frac{\pi}{6}$ is indeed in $[0, \pi]$. Could there be another angle in $[0, \pi]$ with the same cosine? Yes — $\cos(\frac{11\pi}{6})$ also equals $\frac{\sqrt{3}}{2}$, but $\frac{11\pi}{6}$ is not in $[0, \pi]$ (it’s $> \pi$). The only candidate in the principal range is $\frac{\pi}{6}$.

4. Therefore:

Method: Evaluating $\cos^{-1}(\cos\theta)$ for a large angle

Use this when $\theta$ exceeds $2\pi$ (or is very negative) inside $\cos^{-1}(\cos\theta)$.

Steps

Step 1: Reduce $\theta$ modulo $2\pi$ first.

Since cosine has period $2\pi$, subtract whole turns to land in $[0, 2\pi)$:

This strips away the "large" part without changing the cosine.

Step 2: Apply the principal-range logic to the reduced angle.

If the reduced angle lies in $[0,\pi]$, that is the answer. If it lies in $(\pi, 2\pi)$, use $\cos^{-1}(\cos\alpha) = 2\pi - \alpha$.

Step 3: Confirm the result is in $[0,\pi]$.

The final value must sit in the principal range of $\cos^{-1}$; anything outside signals a reduction slip.

Common Mistakes

Mistake 1: Skipping the reduction modulo $2\pi$.

Why it's wrong: $\frac{13\pi}{6}$ is larger than $2\pi$, so working with it directly is error-prone. Correct approach: reduce first, $\frac{13\pi}{6} = 2\pi + \frac{\pi}{6}$, so $\cos\frac{13\pi}{6} = \cos\frac{\pi}{6}$.

Mistake 2: Applying $\cos^{-1}(\cos\theta) = 2\pi - \theta$ to the unreduced angle.

Why it's wrong: that identity is for angles in $(\pi, 2\pi)$, and $\frac{13\pi}{6}$ is not in that interval. Correct approach: after reducing to $\frac{\pi}{6}$, which is already in $[0,\pi]$, the answer is simply $\frac{\pi}{6}$.