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# Find the value of: $\cos^{-1} (\cos \large\frac{13\pi}{6} )$

$\begin{array}{1 1} \frac{\pi}{6} \\ \frac{- \pi}{6} \\ \frac{13 \pi}{6} \\ \frac{-13 \pi}{6} \end{array}$

Toolbox:
• Principal interval of cos is [0,$\pi$]
• cos$(2\pi+x)$=cosx
Since $\large\frac{13\pi}{6}$ is not within the principal interval reduce it within the principal limit $\rightarrow$ $\large\frac{13\pi}{6}=2\pi+\large\frac{\pi}{6}$

The given expression can be written as $cos^{-1}cos \bigg[ 2\pi + \large\frac{\pi}{6} \bigg]$
We know that $cos (2\pi+x)=cosx \rightarrow$ we can represent cos$(2\pi+\large\frac{\pi}{6})$ as $cos\large\frac{\pi}{6}$
$\Rightarrow cos^{-1} cos\large\frac{\pi}{6} = \large\frac{\pi}{6}$

edited Mar 15, 2013