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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} $

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  • 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 \(cos\large\frac{\pi}{6}\)
\( \Rightarrow cos^{-1} cos\large\frac{\pi}{6} = \large\frac{\pi}{6}\)


answered Feb 23, 2013 by thanvigandhi_1
edited Mar 15, 2013 by thanvigandhi_1
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