Answer :
Answer:
The value of [tex]\sin ^ { - 1 } ( \sin 12 ) + \cos ^ { - 1 } ( \cos 12 )[/tex] is equal to 0.
To find:
The value of
[tex]\sin ^ { - 1 } ( \sin 12 ) + \cos ^ { - 1 } ( \cos 12 )[/tex]
Solution:
We know that the principal value of
[tex]\sin ^ { - 1 } x \in \left[ - \frac { \pi } { 2 } , \frac { \pi } { 2 } \right] , \forall x \in [ - 1,1 ][/tex]
and
[tex]\cos ^ { - 1 } x \in [ 0 , \pi ] , \forall x \in [ - 1,1 ][/tex]
As
[tex]12 \notin \left[ \frac { - \pi } { 2 } , \frac { \pi } { 2 } \right] \text { or } 12 \notin [ 0 , \pi ][/tex]
So
[tex]\sin ^ { - 1 } ( \sin 12 ) \neq 12[/tex]
and
[tex]\cos ^ { - 1 } ( \cos 12 ) \neq 12[/tex]
Now,
[tex]12 - 4 \pi \in \left[ - \frac { \pi } { 2 } , \frac { \pi } { 2 } \right][/tex]
and
[tex]4 \pi - 12 \in [ 0 , \pi ][/tex]
So,
[tex]\sin ^ { - 1 } ( \sin 12 ) + \cos ^ { - 1 } ( \cos 12 ) = \sin ^ { - 1 } ( \sin ( 12 - 4 \pi ) ) + \cos ^ { - 1 } ( \cos ( 4 \pi - 12 ) )[/tex]
[tex]\begin{array} { c } { = 12 - 4 \pi + 4 \pi - 12 } \\\\ { = 0 } \end{array}[/tex]
[tex]\sin ^ { - 1 } ( \sin 12 ) + \cos ^ { - 1 } ( \cos 12 ) = 0[/tex]
Thus, the value of [tex]\sin ^ { - 1 } ( \sin 12 ) + \cos ^ { - 1 } ( \cos 12 )[/tex] is equal to 0.