Question

Find the value of the trigonometric expression ${\cos ^{ - 1}}\left( {\cos 12} \right) - {\sin ^{ - 1}}\left( {\sin 12} \right)$.

A.{\text{ }}0 \\\ B.{\text{ }}\pi \\\ C.{\text{ }}8\pi - 24 \\\ $$ $$D.$$ None of these

Answer 1

Hint:

Draw graph of ${\cos ^{ - 1}}\left( {\cos x} \right)$ and ${\text{si}}{{\text{n}}^{ - 1}}\left( {\sin x} \right)$.

Now as we can see that,
We have to find the value of ${\cos ^{ - 1}}\left( {\cos 12} \right)$ and ${\text{si}}{{\text{n}}^{ - 1}}\left( {\sin 12} \right)$ from the above graph.
So, $x$ will be equal to 12 in the above graphs.
Now as we can see from the above graphs of ${\cos ^{ - 1}}\left( {\cos x} \right)$ and ${\text{si}}{{\text{n}}^{ - 1}}\left( {\sin x} \right)$,
That principle range of ${\cos ^{ - 1}}\left( {\cos x} \right)$ and ${\text{si}}{{\text{n}}^{ - 1}}\left( {\sin x} \right)$ is $\left[ {0,2\pi } \right]$.
So, we have to change 12 in terms of $\pi$
Now as we know that, $\pi = 3.14$.
So, $4\pi = 4*(3.14) = 12.56 > 12$
$3\pi = 3*(3.14) = 9.42 < 12$
And, $\dfrac{{7\pi }}{2} = \dfrac{{7*(3.14)}}{2} = 10.99 < 12$
So, $\dfrac{{7\pi }}{2} < 12 < 4\pi$
So, according to the graph drawn above.
Value of ${\cos ^{ - 1}}\left( {\cos 12} \right)$ will be $4\pi - x$, where $x = 12$
And, value of ${\text{si}}{{\text{n}}^{ - 1}}\left( {\sin 12} \right)$ will be $x - 4\pi$, where $x = 12$.
So, ${\cos ^{ - 1}}\left( {\cos 12} \right) = 4\pi - 12$ (1)
And, ${\sin ^{ - 1}}\left( {\sin 12} \right) = 12 - 4\pi$ (2)
Now, subtracting equation 1 and 2. We get,
${\cos ^{ - 1}}\left( {\cos 12} \right) - {\sin ^{ - 1}}\left( {\sin 12} \right) = 8\pi - 24$
Hence, the correct option will be C.

Note:

Whenever we came up with this type of question then we first draw the graph of each trigonometric function. And then find the range in which value of $x$ lies.
After that we can get values of trigonometric functions from the graph. Which can be then manipulated to get the required value of the given equation.