Question
$\cos ^{-1}\left(\cos \frac{7 \pi}{6}\right)$ is equal
✓
Answer
$\cos ^{-1}\left(\cos \frac{7 \pi}{6}\right)$
$($For $\cos^{-1}(\cos x) $ type of problem we have to always check whether the angle is in the principal range or not. This angle must be in the principal range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$)
So here,
$\frac{7 \pi}{6} \notin[0, \pi]$
Now, $\cos ^{-1}\left(\cos \frac{7 \pi}{6}\right)$can be written as,
$\cos ^{-1}\left(\cos \frac{7 \pi}{6}\right)$
$=\cos ^{-1}\left[\cos \left(\pi+\frac{\pi}{6}\right)\right]$
$= -\cos^{-1}$$\left(\cos \frac{\pi}{6}\right)$, where $-\frac{\pi}{6} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right], [$since, $\cos(\pi + x) = -\cos x]$
$= \pi - \cos^{-1} \left(\cos \frac{\pi}{6}\right)$ as $cos^{-1}(-x) = \pi – cos^{-1}$
$= \pi-\frac{\pi}{6}=\frac{5 \pi}{6}$
Hence, $\cos \left(\cos \frac{7 \pi}{6}\right)=\frac{5 \pi}{6}$
$($For $\cos^{-1}(\cos x) $ type of problem we have to always check whether the angle is in the principal range or not. This angle must be in the principal range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$)
So here,
$\frac{7 \pi}{6} \notin[0, \pi]$
Now, $\cos ^{-1}\left(\cos \frac{7 \pi}{6}\right)$can be written as,
$\cos ^{-1}\left(\cos \frac{7 \pi}{6}\right)$
$=\cos ^{-1}\left[\cos \left(\pi+\frac{\pi}{6}\right)\right]$
$= -\cos^{-1}$$\left(\cos \frac{\pi}{6}\right)$, where $-\frac{\pi}{6} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right], [$since, $\cos(\pi + x) = -\cos x]$
$= \pi - \cos^{-1} \left(\cos \frac{\pi}{6}\right)$ as $cos^{-1}(-x) = \pi – cos^{-1}$
$= \pi-\frac{\pi}{6}=\frac{5 \pi}{6}$
Hence, $\cos \left(\cos \frac{7 \pi}{6}\right)=\frac{5 \pi}{6}$
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