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The value of $\cos^{-1}\bigg(\cos\frac{14\pi}{3}\bigg)$ is _____________.

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Toolbox:
  • Principal interval of cos is \([o,\pi]\)
  • \(cos(2n\pi+x)=cosx\) for all \(n\in\:N\)
Ans: \( \large\frac{2\pi}{3}\)
 
Since \(\large\frac{14\pi}{3}\) is not in the principal interval reduce it to its principal interval
\(\large\frac{14\pi}{3}=4\pi+\large\frac{2\pi}{3}\)
 
From the above formula we get
\(\Rightarrow\:cos\large\frac{14\pi}{3}=cos(4\pi+\large\frac{2\pi}{2})=cos\large\frac{2\pi}{3}\)
\(\large\frac{2\pi}{3}\) lies within the principal interval
 
\(\Rightarrow\: cos^{-1} \bigg[ cos \bigg( 4\pi+\large\frac{2\pi}{3} \bigg) \bigg] = cos^{-1}cos \large\frac{2\pi}{3} =\large\frac{2\pi}{3}\)

 

answered Feb 18, 2013 by thanvigandhi_1
edited Mar 16, 2013 by thanvigandhi_1
 
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