Find the principal value of the following:

$\cos^{-1} \left(\cos \frac{13\pi}{6}\right)$

Question

Find the principal value of the following:

$\cos^{-1} \left(\cos \frac{13\pi}{6}\right)$

Accepted Answer

> [!TLDR]
> The principal value of $\cos^{-1}(\cos x)$ is the unique angle in $[0,\pi]$ that has the same cosine as $x$. For $x = \frac{13\pi}{6}$, we reduce it to $\frac{\pi}{6}$ because $\cos(\frac{13\pi}{6}) = \cos(\frac{\pi}{6})$ and $\frac{\pi}{6}$ lies in the principal range $[0,\pi]$. The answer is $\frac{\pi}{6}$.

The key to solving $\cos^{-1}(\cos 	heta)$ 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 $2\pi$.  
   $$
   \frac{13\pi}{6} = 2\pi + \frac{\pi}{6}
   $$
   Since cosine has period $2\pi$,  
   $$
   \cos\left(\frac{13\pi}{6}ight) = \cos\left(2\pi + \frac{\pi}{6}ight) = \cos\left(\frac{\pi}{6}ight) = \frac{\sqrt{3}}{2}.
   $$

2. **Now the problem becomes:**  
   $$
   \cos^{-1}\left(\frac{\sqrt{3}}{2}ight)
   $$
   We need the angle $	heta$ in $[0, \pi]$ such that $\cos	heta = \frac{\sqrt{3}}{2}$.

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 $\frac{\pi}{6}$.

> [!WARNING]
> A common mistake is to cancel $\cos^{-1}$ and $\cos$ directly and write $\frac{13\pi}{6}$. But $\frac{13\pi}{6} \approx 3.93$ radians, which is gre

Inverse function	Domain (allowed inputs $x$ )	Principal value branch (angles returned)
$\sin^{-1} x$	$[-1,1]$	$\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$
$\cos^{-1} x$	$[-1,1]$	$[0,\pi]$
$\tan^{-1} x$	$\mathbb{R}$	$\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$
$\cot^{-1} x$	$\mathbb{R}$	$(0,\pi)$
$\sec^{-1} x$	$(-\infty,-1]\cup[1,\infty)$	$[0,\pi]\setminus\{\tfrac{\pi}{2}\}$
$\csc^{-1} x$	$(-\infty,-1]\cup[1,\infty)$	$\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]\setminus\{0\}$

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

Concept understanding — Principal Value Domain

Principal Value Domain (Principal Branch)

The idea: restrict to one clean interval

Why the intervals differ

Q.Find the principal value of the following: cos⁡−1(cos⁡13π6)\cos^{-1} \left(\cos \frac{13\pi}{6}\right)cos−1(cos613π​)

Concept understanding — Principal Value Domain

Principal Value Domain (Principal Branch)

The idea: restrict to one clean interval

Why the intervals differ

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

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