Question

The value of ${\sin ^{ - 1}}\left( {\sin 12} \right) + {\cos ^{ - 1}}\left( {\cos 12} \right)$ is equal to:
A.Zero
B.$24 - 2\pi $
C.$4\pi - 24$
D.None of these

Answer 1

Hint: We will try to calculate the value of ${\sin ^{ - 1}}\left( {\sin 12} \right)$ and ${\cos ^{ - 1}}\left( {\cos 12} \right)$ separately. Here, the range of ${\sin ^{ - 1}}\left( {\sin 12} \right)$ must lie between $ - \dfrac{\pi }{2}$ and $\dfrac{\pi }{2}$. The range of ${\cos ^{ - 1}}\left( {\cos 12} \right)$ must lie between 0 and $\pi $.

Complete step-by-step answer:
Here the required expression can be broken into two parts. We will calculate the value separately for ${\sin ^{ - 1}}\left( {\sin 12} \right)$and ${\cos ^{ - 1}}\left( {\cos 12} \right)$.
Let us first calculate the value of ${\sin ^{ - 1}}\left( {\sin 12} \right)$.
Let us consider the value of ${\sin ^{ - 1}}\left( {\sin 12} \right)$ to be equal to $\theta $.
Then $\sin \theta = \sin 12$
The general solution for the above equation will be given as $\theta = 2\pi k + 12$, where $k$ is an integer. But the range of $\theta $ must lie between $ - \dfrac{\pi }{2}$ to $\dfrac{\pi }{2}$, as the principle range of the ${\sin ^{ - 1}}\theta $ lies between $ - \dfrac{\pi }{2}$ to $\dfrac{\pi }{2}$. Therefore the value of $k$ must be so chosen that the value of $\theta $ lies between $ - \dfrac{\pi }{2}$ to $\dfrac{\pi }{2}$.
$ - \dfrac{\pi }{2} < 2\pi k + 12$ and
$2\pi k + 12 < \dfrac{\pi }{2}$
and $k$ is an integer.
$ - \dfrac{1}{4} - \dfrac{6}{\pi } < k$
$k \geqslant - 2$ as $k$ is an integer.
Similarly,
$k < - \dfrac{1}{4} - \dfrac{6}{\pi }$
$k \leqslant - 2$
Therefore the value of $k$ is $ - 2$.
And thus $\theta = 2\pi \left( { - 2} \right) + 12$
$\theta = - 4\pi + 12$
We will do the same steps to find the value of ${\cos ^{ - 1}}\left( {\cos 12} \right)$.
Let ${\cos ^{ - 1}}\left( {\cos 12} \right) = \varphi $
Then \[\cos \varphi = \cos 12\]
The general solution for the above equation will be given as \[\varphi = 2\pi k \pm 12\], where \[k\] is an integer. But the range of \[\varphi \] must lie between 0 to \[\pi \], as the principle range of the \[{\cos ^{ - 1}}\varphi \] lies between 0 to \[\pi \]. Therefore the value of \[k\] must be so chosen that the value of \[\varphi \] lies between 0 to \[\pi \].
\[0 < 2\pi k \pm 12\] and
\[2\pi k \pm 12 < \pi \]
and \[k\] is an integer.
\[ \pm \dfrac{6}{\pi } < k\]
\[k \geqslant - 2\]or \[k \geqslant 2\],as \[k\] is an integer.
Similarly,
\[k < \dfrac{1}{2} \pm \dfrac{6}{\pi }\]
\[k \leqslant 2\] or \[k \leqslant - 2\]
Therefore the value of \[k\] satisfying the equations is 2, along with taking a subtraction between 12 and \[2\pi k\].
And thus \[\varphi = 2\pi \left( 2 \right) - 12\]
\[\varphi = 4\pi - 12\]
Adding both values will give us the value for the required expression.
\[ - 4\pi + 4\pi + 12 - 12 = 0\]

Note: The general solution for \[\sin x = \sin \theta \] is given by \[x = 2\pi k + \theta \], where \[k\] is an integer and the range of \[x\] lies between \[ - \dfrac{\pi }{2}\] and \[\dfrac{\pi }{2}\]. Similarly, the general solution for \[\cos x = \cos \theta \] is given by \[x = 2\pi k \pm \theta \], where \[k\] is an integer and the range of \[x\] lies between 0 and \[\pi \].