Question

The value of ${sin}^{-1}\left(sin12\right)+{cos}^{-1}\left(cos12\right)$ is equal to

• A. zero
• B. $24-4\pi$
• C. $4\pi -24$
• D. none of these

Answer 1

The correct option is A zero

Pricipal value of ${\sin }^{-1}x\in [-\frac{\pi }{2},\frac{\pi }{2}],\forall x\in [-1,1]$

& Principal value of $co{s}^{-1}p\in [0,\pi ],\forall p\in [-1,1]$

$\therefore {\sin }^{-1}\left(\sin 12\right)\ne 12\notin [-\frac{\pi }{2},\frac{\pi }{2}]$

and

$\cos -1\left(\cos 12\right)\ne 12\notin [0,\pi ].$

So,

$si{n}^{-1}\left(\sin 12\right)+co{s}^{-1}\left(\cos 12\right)$

$=si{n}^{-1}\left(\sin \right(12-4\pi \left)\right)+co{s}^{-1}\left(\cos \right(4\pi -12\left)\right)$

$=(12-4\pi )+(4\pi -12)$

$\therefore si{n}^{-1}\left(\sin 12\right)+co{s}^{-1}\left(\cos 12\right)=0$