Question

The value of ${\cos }^{-1}\left(\cos 12\right)-{\sin }^{-1}\left(\sin 12\right)$ is

• A. $0$
• B. $\pi$
• C. $8\pi -24$
• D. none of these

Answer 1

The correct option is C $8\pi -24$

$12rad$ lies in $4th$ quadrant
$\frac{7\pi }{2}<12<4\pi$
Let $\theta$ be an acute angle such that
$12+\theta =4\pi$
$\therefore 12=4\pi -\theta$ or $\theta =4\pi -12$
${\cos }^{-1}\left(\cos 12\right)-{\sin }^{-1}\left(\sin 12\right)$
$={\cos }^{-1}\left(\cos \right(4\pi -\theta \left)\right)-{\sin }^{-1}\left(\sin \right(4\pi -\theta \left)\right)$
$={\cos }^{-1}\left(\cos \theta \right)-{\sin }^{-1}(-\sin \theta )$
$={\cos }^{-1}\left(\cos \theta \right)-{\sin }^{-1}\left(\sin \right(-\theta \left)\right)$
$=\theta -(-\theta )$
$=2\theta$
$=2(4\pi -12)$
$=8\pi -24$
$\therefore {\cos }^{-1}\left(\cos 12\right)-{\sin }^{-1}\left(\sin 12\right)=8\pi -24$