Q.The value of $\cos^{-1}\left(\cos\frac{14\pi}{3}\right)$ is __________.

Concept: Range-restricted inverse cosine — $\cos^{-1}(\cos x)$ returns the principal value in $[0,\pi]$, so we must reduce the angle to this interval.

Step 1: Reduce $\frac{14\pi}{3}$ modulo $2\pi$ to find a coterminal angle.

$\frac{14\pi}{3} = 4\pi + \frac{2\pi}{3}$ (since $4\pi = \frac{12\pi}{3}$). So the angle is $\frac{2\pi}{3}$. …

The function $\cos^{-1}(\cos x)$ is not simply $x$ — it gives the angle in the principal range $[0,\pi]$ that has the same cosine as $x$. Think of it as a "wrapper" that folds any angle into this interval. So when you see $\cos^{-1}(\cos \theta)$, the first step is always to bring $\theta$ into a manageable form using the periodicity of cosine: $\cos(\theta + 2\pi n) = \cos \theta$ for any integer $n$.

Here, $\frac{14\pi}{3}$ is large — let's see where it lands.

1. Reduce modulo $2\pi$.

   Compute $\frac{14\pi}{3} \div 2\pi = \frac{14\pi}{3} \cdot \frac{1}{2\pi} = \frac{7}{3} = 2 + \frac{1}{3}$.

   So $\frac{14\pi}{3} = 2 \cdot 2\pi + \frac{2\pi}{3}$.

   Since $\cos$ has period $2\pi$, $\cos\left(\frac{14\pi}{3}\right) = \cos\left(\frac{2\pi}{3}\right)$.

2. Check the principal range.

   The principal value of $\cos^{-1}$ is defined to lie in $[0,\pi]$.

   $\frac{2\pi}{3}$ is approximately $2.094$ radians, which is indeed in $[0,\pi]$ (since $\pi \approx 3.14$).

   So $\cos^{-1}\left(\cos\frac{2\pi}{3}\right) = \frac{2\pi}{3}$. …

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

Steps

Step 1: Reduce $x$ modulo $2\pi$.

Use $\cos(x+2\pi n)=\cos x$ to strip whole turns and land in $[0,2\pi)$.

Step 2: Map the reduced angle into $[0,\pi]$.

• If it is already in $[0,\pi]$, that is the answer. …

Common Mistakes

Mistake 1: Answering $\frac{14\pi}{3}$.

Why it's wrong: $\cos^{-1}$ can only return an angle in $[0,\pi]$. Correct approach: subtract $4\pi$ first to reach $\frac{2\pi}{3}$.

Mistake 2: Reducing modulo $\pi$ instead of $2\pi$. …