Browse Questions

# Find the value of the expression $cos^{-1} \bigg(cos \frac{\large 7\pi}{\large 6} \bigg)$

$\begin{array} (A) \frac{7\pi}{6} \quad & (B) \frac{5\pi}{6} \quad & (C) \frac{\pi}{3} \quad & (D) \frac{\pi}{6} \end{array}$

Toolbox:
• cos($\pi+x)=-cosx$
• $cos(\pi+x)=cos(\pi-x)$
• Principal interval of cos is [0,$\pi$]
Given $cos^{-1} \bigg(cos \frac{\large 7\pi}{\large 6} \bigg)$:
$\frac{7\pi}{6} = \pi+\frac{\pi}{6},\;$ However, $\pi+\frac{\pi}{6}$ is not in the principal interval.
We know that $cos(\pi+x)=cos(\pi-x)$. Therefore by taking $x=$$\frac{\pi}{6}$ we can write
$cos(\pi+\frac{\pi}{6})=cos(\pi-\frac{\pi}{6})$
The expression then reduces to $cos^{-1} \bigg[ cos \bigg( \pi-\frac{\pi}{6} \bigg) \bigg] = cos^{-1}cos \frac{5\pi}{6}$
$= \frac{5\pi}{6}$
edited Mar 14, 2013