Question

What is the principal value of ${{\cos }^{-1}}\left( \cos \dfrac{2\pi }{3} \right)+{{\sin }^{-1}}\left( \sin \dfrac{2\pi }{3} \right)$ ?

Answer 1

Hint: For solving this question first, we will go through some important aspects like domain and range of the inverse trigonometric functions $y={{\cos }^{-1}}x$ and $y={{\sin }^{-1}}x$ . First, we will use one of the basic formulas of the trigonometric ratio to write $\cos \dfrac{2\pi }{3}=-\dfrac{1}{2}$ in the given term. After that, we will use one of the basic formula of inverse trigonometric functions, i.e. ${{\cos }^{-1}}\left( -\dfrac{1}{2} \right)=\dfrac{2\pi }{3}$ to find the value of ${{\cos }^{-1}}\left( \cos \dfrac{2\pi }{3} \right)$ . After that, we will use formula $\sin \dfrac{2\pi }{3}=\dfrac{\sqrt{3}}{2}$ and ${{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)=\dfrac{\pi }{3}$ to find the value of ${{\sin }^{-1}}\left( \sin \dfrac{2\pi }{3} \right)$ . Then, we will easily find the value of ${{\cos }^{-1}}\left( \cos \dfrac{2\pi }{3} \right)+{{\sin }^{-1}}\left( \sin \dfrac{2\pi }{3} \right)$ .

Complete step-by-step solution -
Given:
We have to find the principal value of the following:
${{\cos }^{-1}}\left( \cos \dfrac{2\pi }{3} \right)+{{\sin }^{-1}}\left( \sin \dfrac{2\pi }{3} \right)$
Now, we will find the principal values of ${{\cos }^{-1}}\left( \cos \dfrac{2\pi }{3} \right)$ and ${{\sin }^{-1}}\left( \sin \dfrac{2\pi }{3} \right)$ separately and then, we will add them to find the principal value of ${{\cos }^{-1}}\left( \cos \dfrac{2\pi }{3} \right)+{{\sin }^{-1}}\left( \sin \dfrac{2\pi }{3} \right)$ .
Calculation for the principal value of ${{\cos }^{-1}}\left( \cos \dfrac{2\pi }{3} \right)$ :
Now, before we proceed we should know about the inverse trigonometric function $y={{\cos }^{-1}}x$ . For more clarity look at the figure given below:

In the above figure, the plot $y=f\left( x \right)={{\cos }^{-1}}x$ is shown. And we should know that the function $y={{\cos }^{-1}}x$ is defined for $x\in \left[ -1,1 \right]$ and its range is $y\in \left[ 0,\pi \right]$ then, $y$ is the principal value of ${{\cos }^{-1}}x$ .
Now, we will use the above concept for giving the correct principal value of ${{\cos }^{-1}}\left( \cos \dfrac{2\pi }{3} \right)$ .
Now, before we proceed further we should know the following formulas:
$\begin{align}
  & \cos \dfrac{2\pi }{3}=\cos \left( \pi -\dfrac{\pi }{3} \right)=-\cos \dfrac{\pi }{3}=-\dfrac{1}{2}..................\left( 1 \right) \\
 & {{\cos }^{-1}}\left( -\dfrac{1}{2} \right)=\dfrac{2\pi }{3}...........\left( 2 \right) \\
\end{align}$
Now, we will use the above two formulas to solve this question.
We have, ${{\cos }^{-1}}\left( \cos \dfrac{2\pi }{3} \right)$ .
Now, we will use the formula from the equation (1) to write $\cos \dfrac{2\pi }{3}=-\dfrac{1}{2}$ in the term ${{\cos }^{-1}}\left( \cos \dfrac{2\pi }{3} \right)$ . Then,
$\begin{align}
  & {{\cos }^{-1}}\left( \cos \dfrac{2\pi }{3} \right) \\
 & \Rightarrow {{\cos }^{-1}}\left( -\dfrac{1}{2} \right) \\
\end{align}$
Now, we will use the formula from the equation (2) to write ${{\cos }^{-1}}\left( -\dfrac{1}{2} \right)=\dfrac{2\pi }{3}$ in the above line. Then,
$\begin{align}
  & {{\cos }^{-1}}\left( -\dfrac{1}{2} \right) \\
 & \Rightarrow \dfrac{2\pi }{3} \\
\end{align}$
Now, from the above result, we conclude that the principal value of the expression ${{\cos }^{-1}}\left( \cos \dfrac{2\pi }{3} \right)$ will be equal to $\dfrac{2\pi }{3}$ . Then,
${{\cos }^{-1}}\left( \cos \dfrac{2\pi }{3} \right)=\dfrac{2\pi }{3}.....................\left( 3 \right)$
Calculation for the principal value of ${{\sin }^{-1}}\left( \sin \dfrac{2\pi }{3} \right)$ :
Now, before we proceed we should know about the inverse trigonometric function $y={{\sin }^{-1}}x$ . For more clarity look at the figure given below:

In the above figure, the plot $y=f\left( x \right)={{\sin }^{-1}}x$ is shown. And we should know that the function $y={{\sin }^{-1}}x$ is defined for $x\in \left[ -1,1 \right]$ and its range is $y\in \left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$ then, $y$ is the principal value of ${{\sin }^{-1}}x$ .
Now, we will use the above concept for giving the correct principal value of ${{\sin }^{-1}}\left( \sin \dfrac{2\pi }{3} \right)$ .
Now, before we proceed further we should know the following formulas:
$\begin{align}
  & \sin \dfrac{2\pi }{3}=\sin \left( \pi -\dfrac{\pi }{3} \right)=\sin \dfrac{\pi }{3}=\dfrac{\sqrt{3}}{2}.................\left( 4 \right) \\
 & {{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)=\dfrac{\pi }{3}...........\left( 5 \right) \\
\end{align}$
Now, we will use the above two formulas to solve this question.
We have, ${{\sin }^{-1}}\left( \sin \dfrac{2\pi }{3} \right)$ .
Now, we will use the formula from the equation (4) to write $\sin \dfrac{2\pi }{3}=\dfrac{\sqrt{3}}{2}$ in the term ${{\sin }^{-1}}\left( \sin \dfrac{2\pi }{3} \right)$ . Then,
$\begin{align}
  & {{\sin }^{-1}}\left( \sin \dfrac{2\pi }{3} \right) \\
 & \Rightarrow {{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right) \\
\end{align}$
Now, we will use the formula from the equation (5) to write ${{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)=\dfrac{\pi }{3}$ in the above line. Then,
$\begin{align}
  & {{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right) \\
 & \Rightarrow \dfrac{\pi }{3} \\
\end{align}$
Now, from the above result, we conclude that the principal value of the expression ${{\sin }^{-1}}\left( \sin \dfrac{2\pi }{3} \right)$ will be equal to $\dfrac{\pi }{3}$ . Then,
${{\sin }^{-1}}\left( \sin \dfrac{2\pi }{3} \right)=\dfrac{\pi }{3}.....................\left( 6 \right)$
Now, we will use the result of the equation (3) and (6) to find the principal value of ${{\cos }^{-1}}\left( \cos \dfrac{2\pi }{3} \right)+{{\sin }^{-1}}\left( \sin \dfrac{2\pi }{3} \right)$ . Then,
$\begin{align}
  & {{\cos }^{-1}}\left( \cos \dfrac{2\pi }{3} \right)+{{\sin }^{-1}}\left( \sin \dfrac{2\pi }{3} \right) \\
 & \Rightarrow \dfrac{2\pi }{3}+\dfrac{\pi }{3} \\
 & \Rightarrow \pi \\
\end{align}$
Now, from the above result, we conclude that principal value of the ${{\cos }^{-1}}\left( \cos \dfrac{2\pi }{3} \right)+{{\sin }^{-1}}\left( \sin \dfrac{2\pi }{3} \right)$ is equal to $\pi $ .
Thus, ${{\cos }^{-1}}\left( \cos \dfrac{2\pi }{3} \right)+{{\sin }^{-1}}\left( \sin \dfrac{2\pi }{3} \right)=\pi $.

Note: Here, the student should first understand what is asked in the question and then proceed in the right direction to get the correct answer quickly. Moreover, we should avoid writing ${{\sin }^{-1}}\left( \sin \dfrac{2\pi }{3} \right)=\dfrac{2\pi }{3}$ directly and use the basic concepts of domain and range of the inverse trigonometric functions $y={{\cos }^{-1}}x$ and $y={{\sin }^{-1}}x$ correctly. And after giving calculating the values of ${{\cos }^{-1}}\left( \cos \dfrac{2\pi }{3} \right)$ and ${{\sin }^{-1}}\left( \sin \dfrac{2\pi }{3} \right)$ , we should check for the validity of our answer by checking whether it lies in the range of its inverse trigonometric function or not.