Question

Find the principal value of ${{\cos }^{-1}}\left( \dfrac{1}{2} \right)-2{{\sin }^{-1}}\left( -\dfrac{1}{2} \right)$?

Answer 1

We start solving the problem by assigning a variable ‘x’ to the value of ${{\cos }^{-1}}\left( \dfrac{1}{2} \right)-2{{\sin }^{-1}}\left( -\dfrac{1}{2} \right)$. We first find the principal values of ${{\cos }^{-1}}\left( \dfrac{1}{2} \right)$ and ${{\sin }^{-1}}\left( -\dfrac{1}{2} \right)$ by following the principal range of ${{\cos }^{-1}}\left( x \right)$ and ${{\sin }^{-1}}\left( x \right)$. After finding the values, we use them and calculate to get the required value ‘x’.

Complete step by step answer:
According to the problem, we need to find the principal value of ${{\cos }^{-1}}\left( \dfrac{1}{2} \right)-2{{\sin }^{-1}}\left( -\dfrac{1}{2} \right)$.
Let us assume the principal value of ${{\cos }^{-1}}\left( \dfrac{1}{2} \right)-2{{\sin }^{-1}}\left( -\dfrac{1}{2} \right)$ be ’x’.
So, we have got $x={{\cos }^{-1}}\left( \dfrac{1}{2} \right)-2{{\sin }^{-1}}\left( -\dfrac{1}{2} \right)$.
Let us find principal values of ${{\cos }^{-1}}\left( \dfrac{1}{2} \right)$ and ${{\sin }^{-1}}\left( -\dfrac{1}{2} \right)$ first.
We know that the principal range of ${{\cos }^{-1}}x$ is $\left[ 0,\pi \right]$. We know that the value of $\cos \left( \dfrac{\pi }{3} \right)$ is $\dfrac{1}{2}$.
So, we have got the principal value ${{\cos }^{-1}}\left( \dfrac{1}{2} \right)={{\cos }^{-1}}\left( \cos \left( \dfrac{\pi }{3} \right) \right)$ ---(1).
We know that ${{\cos }^{-1}}\left( \cos x \right)=x$, if the value of ‘x’ lies in the interval $\left[ 0,\pi \right]$. We use this result in equation (1).
So, we have got the principal value ${{\cos }^{-1}}\left( \dfrac{1}{2} \right)=\dfrac{\pi }{3}$ ---(2).
Now, we find the principal value of ${{\sin }^{-1}}\left( -\dfrac{1}{2} \right)$.
We know that the principal range of ${{\sin }^{-1}}x$ is $\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]$. We know that the value of $\sin \left( \dfrac{-\pi }{6} \right)$ is $\dfrac{-1}{2}$.
So, we have got the principal value ${{\sin }^{-1}}\left( \dfrac{-1}{2} \right)={{\sin }^{-1}}\left( \sin \left( \dfrac{-\pi }{6} \right) \right)$ ---(3).
We know that ${{\sin }^{-1}}\left( \sin x \right)=x$, if the value of ‘x’ lies in the interval $\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]$. We use this result in equation (3).
So, we have got the principal value ${{\sin }^{-1}}\left( \dfrac{-1}{2} \right)=\dfrac{-\pi }{6}$ ---(4).
Let us now find the principal value of ‘x’.
We have got the principal value of $x={{\cos }^{-1}}\left( \dfrac{1}{2} \right)-2{{\sin }^{-1}}\left( -\dfrac{1}{2} \right)$.
From equations (3) and (4), we have got the principal value of $x=\dfrac{\pi }{3}-2\left( \dfrac{-\pi }{6} \right)$.
We have got the principal value of $x=\dfrac{\pi }{3}-\left( \dfrac{-\pi }{3} \right)$.
We have got the principal value of $x=\dfrac{\pi }{3}+\dfrac{\pi }{3}$.
We have got the principal value of $x=\dfrac{2\pi }{3}$.
We have found the principal value of ${{\cos }^{-1}}\left( \dfrac{1}{2} \right)-2{{\sin }^{-1}}\left( -\dfrac{1}{2} \right)$ as $\dfrac{2\pi }{3}$.

∴ The principal value of ${{\cos }^{-1}}\left( \dfrac{1}{2} \right)-2{{\sin }^{-1}}\left( -\dfrac{1}{2} \right)$ is $\dfrac{2\pi }{3}$.

Note: We should not confuse the principal range of ${{\cos }^{-1}}\left( x \right)$ with $\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]$ as the values of $\cos x$ lies in $\left[ 0,1 \right]$ within that interval. If the principal value is not mentioned in the problem, we could have got more than one solution. Similarly, we can get problems finding the solution set for x (the assigned value in the beginning).