Browse Questions

# The value of $\sin^{-1}\bigg(\cos\frac{33\pi}{5}\bigg)$ is

$\begin{array}{1 1} \frac{3 \pi}{5} \\ \frac{-7 \pi }{5} \\ \frac{\pi}{10} \\ \frac{-\pi}{10} \end{array}$

Toolbox:
• $cos(2n\pi+\theta)=cos\theta$
• Principal interval of cos is [0,$\pi$]
• $cosx=sin(\frac{\pi}{2}-x)$
• Principal interval of sin is $[-\frac{\pi}{2},\frac{\pi}{2}]$
Ans (D) $\frac{-\pi}{10}$
Reduce the angle $\frac{33\pi}{5}$into the principal interval of cos
$\frac{33\pi}{5}=6\pi + \frac{3\pi}{5}$
$sin^{-1} \bigg( cos \bigg( \frac{33\pi}{5} \bigg) \bigg)$
$= sin^{-1} \bigg[ cos \bigg[ 6\pi + \frac{3\pi}{5} \bigg] \bigg]$
By taking $\theta=\frac{3\pi}{5}$, and n=3 in the above formula, we get
$cos \big( 6\pi + \frac{3\pi}{5}\big)=cos\frac{3\pi}{5}$
$\Rightarrow\: sin^{-1}cos \frac{3\pi}{5}$
$= sin^{-1}sin \big[ \frac{\pi}{2}-\frac{3\pi}{5} \big] =\frac{\pi}{2}-\frac{3\pi}{5}$
$= -\frac{\pi}{10}$
edited Mar 8, 2013