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

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

$\Rightarrow\: cos^{-1} \bigg[ cos \bigg( 4\pi+\large\frac{2\pi}{3} \bigg) \bigg] = cos^{-1}cos \large\frac{2\pi}{3} =\large\frac{2\pi}{3}$

edited Mar 16, 2013