Question

The value of ${\cos ^{ - 1}}\left( {\cos \dfrac{{5\pi }}{3}} \right) + {\sin ^{ - 1}}\left( {\cos \dfrac{{5\pi }}{3}} \right)$ is
A.$\dfrac{\pi }{2}$
B.$\dfrac{{5\pi }}{3}$
C.$\dfrac{{10\pi }}{3}$
D.0

Answer 1

Hint: We will first simplify the given expression using the properties of inverse trigonometry such as ${\cos ^{ - 1}}\left( {\cos x} \right) = x$, when $x \in \left[ {0,\pi } \right]$ and ${\sin ^{ - 1}}\left( {\sin x} \right) = x$ when $x \in \left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$. Then, we will substitute and simplify the values to get the required answer.

Complete step by step answer:

We have to find the value of ${\cos ^{ - 1}}\left( {\cos \dfrac{{5\pi }}{3}} \right) + {\sin ^{ - 1}}\left( {\cos \dfrac{{5\pi }}{3}} \right)$
It is known that ${\cos ^{ - 1}}\left( {\cos x} \right) = x$, when $x \in \left[ {0,\pi } \right]$ and ${\sin ^{ - 1}}\left( {\sin x} \right) = x$ when $x \in \left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$
We will rewrite the angles using the formulas of trigonometry.
We can rewrite the angle as\[\cos \left( {\dfrac{{5\pi }}{3}} \right) = \cos \left( {2\pi - \dfrac{\pi }{3}} \right)\]
Also, $\cos \left( {2\pi - \theta } \right) = \cos \theta $
Therefore, \[\cos \left( {2\pi - \dfrac{\pi }{3}} \right) = \cos \dfrac{\pi }{3}\]
On substituting the values in the given expression, we get,
${\cos ^{ - 1}}\left( {\cos \dfrac{\pi }{3}} \right) + {\sin ^{ - 1}}\left( {\cos \dfrac{\pi }{3}} \right)$
Now, we know that $\cos \theta = \sin \left( {\dfrac{\pi }{2} - \theta } \right)$
Hence,
$
  \cos \dfrac{\pi }{3} = \sin \left( {\dfrac{\pi }{2} - \dfrac{\pi }{3}} \right) \\
   \Rightarrow \cos \dfrac{\pi }{3} = \sin \left( {\dfrac{\pi }{6}} \right) \\
$
We can now write the given expression as,
${\cos ^{ - 1}}\left( {\cos \dfrac{\pi }{3}} \right) + {\sin ^{ - 1}}\left( {\sin \dfrac{\pi }{6}} \right)$
We will use the property ${\cos ^{ - 1}}\left( {\cos x} \right) = x$, when $x \in \left[ {0,\pi } \right]$ and ${\sin ^{ - 1}}\left( {\sin x} \right) = x$ when $x \in \left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$ to further simplify.
$\Rightarrow\dfrac{\pi }{3} + \dfrac{\pi }{6} = \dfrac{\pi }{2}$
Hence, option A is correct.


Note: We cannot directly write ${\cos ^{ - 1}}\left( {\cos \dfrac{{5\pi }}{3}} \right)$ as $\dfrac{{5\pi }}{3}$ because the range of ${\cos ^{ - 1}}x$ is $\left[ {0,\pi } \right]$ and $\dfrac{{5\pi }}{3} > \pi $. Similarly, the range of ${\sin ^{ - 1}}x$ is $\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$. Many students make mistakes by simply writing the values without considering the range.