Question

The value of sin inverse of (sin 12) + cos inverse of (cos 12) is equal to 
A) 0            B) 24-2pi              C) 4pi-24             D) 2pi-24
Given the answer is option A please explain me how 

Answer 1

Answer:

The value of $\sin ^ { - 1 } ( \sin 12 ) + \cos ^ { - 1 } ( \cos 12 )$ is equal to 0.

To find:

The value of  

$\sin ^ { - 1 } ( \sin 12 ) + \cos ^ { - 1 } ( \cos 12 )$

Solution:

We know that the principal value of  

$\sin ^ { - 1 } x \in \left[ - \frac { \pi } { 2 } , \frac { \pi } { 2 } \right] , \forall x \in [ - 1,1 ]$

and

$\cos ^ { - 1 } x \in [ 0 , \pi ] , \forall x \in [ - 1,1 ]$

As

$12 \notin \left[ \frac { - \pi } { 2 } , \frac { \pi } { 2 } \right] \text { or } 12 \notin [ 0 , \pi ]$

So

$\sin ^ { - 1 } ( \sin 12 ) \neq 12$

and

$\cos ^ { - 1 } ( \cos 12 ) \neq 12$

Now,

$12 - 4 \pi \in \left[ - \frac { \pi } { 2 } , \frac { \pi } { 2 } \right]$

and

$4 \pi - 12 \in [ 0 , \pi ]$

So,

$\sin ^ { - 1 } ( \sin 12 ) + \cos ^ { - 1 } ( \cos 12 ) = \sin ^ { - 1 } ( \sin ( 12 - 4 \pi ) ) + \cos ^ { - 1 } ( \cos ( 4 \pi - 12 ) )$

$\begin{array} { c } { = 12 - 4 \pi + 4 \pi - 12 } \\\\ { = 0 } \end{array}$

$\sin ^ { - 1 } ( \sin 12 ) + \cos ^ { - 1 } ( \cos 12 ) = 0$

Thus, the value of $\sin ^ { - 1 } ( \sin 12 ) + \cos ^ { - 1 } ( \cos 12 )$ is equal to 0.