Question

Evaluate: ${\cos ^{ - 1}}\left( {\cos \dfrac{{2\pi }}
{3}} \right) + {\sin ^{ - 1}}\left( {\sin \dfrac{{2\pi }} {3}} \right)$
(A) $\dfrac{{4\pi }}
{3}$
(B) $\dfrac{\pi }
{2}$
(C) $\dfrac{{3\pi }}
{4}$
(D) $\pi $

Answer 1

Hint:The given expressions can be evaluated by using the properties of inverse trigonometric functions.
According to property of inverse trigonometric functions, if $x \in \left[ {0,\pi } \right]$; the value of ${\sin ^{ - 1}}\left( {\sin x} \right)$ is $x$.
Also, if $x \in \left[ { - \dfrac{\pi }
{2},\dfrac{\pi }
{2}} \right]$;
the value of ${\cos ^{ - 1}}\left( {\cos x} \right)$ is $x$.

Complete step by step solution:
The given expression is ${\cos ^{ - 1}}\left( {\cos \dfrac{{2\pi }}
{3}} \right) + {\sin ^{ - 1}}\left( {\sin \dfrac{{2\pi }} {3}} \right)$.
In order to evaluate the above expression, we need to solve each of the terms by using the properties of inverse trigonometric functions.
It is known that if $x \in \left[ {0,\pi } \right]$;
the value of ${\sin ^{ - 1}}\left( {\sin x} \right)$ is $x$.
Also, if $x \in \left[ { - \dfrac{\pi }
{2},\dfrac{\pi }
{2}} \right]$;
the value of ${\cos ^{ - 1}}\left( {\cos x} \right)$is $x$.
So, ${\sin ^{ - 1}}\left( {\sin x} \right) = x,\forall x \in \left[ { - \dfrac{\pi }
{2},\dfrac{\pi }
{2}} \right]$ and ${\cos ^{ - 1}}\left( {\cos x} \right) = x;\forall x \in \left[ {0,\pi } \right]$.
Now, simplify the term ${\cos ^{ - 1}}\left( {\cos \dfrac{{2\pi }} {3}} \right)$.
In the above expression, the angle is $\dfrac{{2\pi }} {3}$
which belongs to the interval $\left[ { - \dfrac{\pi }
{2},\dfrac{\pi }
{2}} \right]$.
So, the value of the expression ${\cos ^{ - 1}}\left( {\cos \dfrac{{2\pi }}
{3}} \right)$ is $\dfrac{{2\pi }} {3}$ that is, ${\cos ^{ - 1}}\left( {\cos \dfrac{{2\pi }}
{3}} \right) = \dfrac{{2\pi }}
{3}$.
Also, simplify the term ${\sin ^{ - 1}}\left( {\sin \dfrac{{2\pi }}
{3}} \right)$.
In the above expression, the angle is $\dfrac{{2\pi }}
{3}$ which does not belong to the interval $\left[ { - \dfrac{\pi }
{2},\dfrac{\pi }
{2}} \right]$.
So, rewrite the angle $\dfrac{{2\pi }}
{3}$ as,
$
\dfrac{{2\pi }}
{3} = \dfrac{{3\pi - \pi }}
{3} \\
= \dfrac{{3\pi }}
{3} - \dfrac{\pi }
{3} \\
= \pi - \dfrac{\pi }
{3} \\
 $
Substitute $\pi - \dfrac{\pi }
{3}$
for $\dfrac{{2\pi }}
{3}$
in the expression ${\sin ^{ - 1}}\left( {\sin \dfrac{{2\pi }}
{3}} \right)$
${\sin ^{ - 1}}\left( {\sin \dfrac{{2\pi }}
{3}} \right) = {\sin ^{ - 1}}\left( {\sin \left( {\pi - \dfrac{\pi }
{3}} \right)} \right)$
Also, it is known that $\sin \left( {\pi - \dfrac{\pi }
{3}} \right) = \sin \dfrac{\pi }
{3}$
Substitute $\sin \dfrac{\pi }
{3}$ for $\sin \left( {\pi - \dfrac{\pi }
{3}} \right)$ in the ${\sin ^{ - 1}}\left( {\sin \dfrac{{2\pi }}
{3}} \right) = {\sin ^{ - 1}}\left( {\sin \left( {\pi - \dfrac{\pi }
{3}} \right)} \right)$
$
{\sin ^{ - 1}}\left( {\sin \dfrac{{2\pi }}
{3}} \right) = {\sin ^{ - 1}}\left( {\sin \left( {\pi - \dfrac{\pi }
{3}} \right)} \right) \\
{\sin ^{ - 1}}\left( {\sin \dfrac{{2\pi }}
{3}} \right) = {\sin ^{ - 1}}\left( {\sin \dfrac{\pi }
{3}} \right) \\
 $
In the above expression, it is seen that the angle is $\dfrac{\pi } {3}$ which belongs to the interval $\left[ { - \dfrac{\pi }
{2},\dfrac{\pi }
{2}} \right]$
Rewrite the expression ${\sin ^{ - 1}}\left( {\sin \dfrac{\pi }
{3}} \right)$ by using the property ${\sin ^{ - 1}}\left( {\sin x} \right) = x$
${\sin ^{ - 1}}\left( {\sin \dfrac{\pi }
{3}} \right) = \dfrac{\pi }
{3}$
Substitute the values $\dfrac{{2\pi }}
{3}$
for ${\cos ^{ - 1}}\left( {\cos \dfrac{{2\pi }}
{3}} \right)$
and $\dfrac{\pi }
{3}$
for ${\sin ^{ - 1}}\left( {\sin \dfrac{\pi }
{3}} \right)$
in the expression ${\cos ^{ - 1}}\left( {\cos \dfrac{{2\pi }}
{3}} \right) + {\sin ^{ - 1}}\left( {\sin \dfrac{{2\pi }}
{3}} \right)$
and simplify.
$
{\cos ^{ - 1}}\left( {\cos \dfrac{{2\pi }}
{3}} \right) + {\sin ^{ - 1}}\left( {\sin \dfrac{{2\pi }}
{3}} \right) = \dfrac{{2\pi }}
{3} + \dfrac{\pi }
{3} \\
= \dfrac{{2\pi + \pi }}
{3} \\
= \dfrac{{3\pi }}
{3} \\
= \pi \\
 $
Thus, the required value of the given expression is $\pi $ .
Hence, the required correct answer is (D).

Note: To evaluate the given expression it is necessary to use the properties of inverse trigonometric functions for ${\cos ^{ - 1}}$ and ${\sin ^{ - 1}}$ . We must know that the inverse trigonometric function ${\cos ^{ - 1}}$ lies in the interval $\left[ {0,\pi } \right]$ . Also, the inverse trigonometric function ${\sin ^{ - 1}}$ lies in the interval $\left[ { - \dfrac{\pi } {2},\dfrac{\pi } {2}} \right]$