Question

The principal value of ${{\cos }^{-1}}\left( -\dfrac{1}{2} \right)$ is equal to
(A) $\dfrac{\pi }{3}$
(B)$\dfrac{2\pi }{3}$
(C)$-\dfrac{\pi }{3}$
(D)$-\dfrac{2\pi }{3}$

Answer 1

To solve the question of this type we need to know the concept of inverse trigonometric function. The argument or the domain should be between -1 and +1 which means $-1\le a\le 1$ in case of inverse trigonometric function $\cos$. To make calculation easy the first step is to apply the formula ${{\cos }^{-1}}\left( -a \right)=\pi -{{\cos }^{-1}}\left( a \right)$, where $a$ is the argument.

Complete step-by-step solution:
The question asks us to find the principal value for a trigonometric function $\cos$with the value to be $\left( -\dfrac{1}{2} \right)$. The range of principal value of ${{\cos }^{-1}}$ is $\left[ 0,\pi \right]$ The presence of square brackets means the lower and the upper limit is included. So in this case $0$ and $\pi$ are included in the range of ${{\cos }^{-1}}$ .
Since the argument given in the question is a negative integer, so to make it a bit easy we will change the argument of the inverse trigonometric function positive. To do this we will use the formula
${{\cos }^{-1}}\left( -a \right)=\pi -{{\cos }^{-1}}\left( a \right)$
Here $a$ varies as $0\le a\le 1$ .
Now applying the same formula with the argument given to us we get:
$\Rightarrow {{\cos }^{-1}}\left( -\dfrac{1}{2} \right)=\pi -{{\cos }^{-1}}\left( \dfrac{1}{2} \right)$
On checking the value of ${{\cos }^{-1}}\left( \dfrac{1}{2} \right)$ the angle we get $\dfrac{\pi }{3}$. Substituting the values in the formula:
$\Rightarrow {{\cos }^{-1}}\left( -\dfrac{1}{2} \right)=\pi -\dfrac{\pi }{3}$
Taking $3$ as L.C.M for the number on R.H.S. we get:
$\Rightarrow {{\cos }^{-1}}\left( -\dfrac{1}{2} \right)=\dfrac{3\pi -\pi }{3}$
$\Rightarrow {{\cos }^{-1}}\left( -\dfrac{1}{2} \right)=\dfrac{2\pi }{3}$
$\therefore$ The principal value of ${{\cos }^{-1}}\left( -\dfrac{1}{2} \right)$ is equal to $\dfrac{2\pi }{3}$.

Note: We can check whether the answer we got is correct or not. To check this we will find the value of $\cos \dfrac{2\pi }{3}$. To solve we will apply the formula:
$\cos \theta =-\cos \left( \pi -\theta \right)$
The negative sign shows that the value of the trigonometric function $\cos$ is negative when $\dfrac{\pi }{2}<\theta <\pi$ that means when the angle is at second quadrant.
$\Rightarrow \cos \dfrac{2\pi }{3}=-\cos \left( \pi -\dfrac{2\pi }{3} \right)$
$\Rightarrow \cos \dfrac{2\pi }{3}=-\cos \left( \dfrac{\pi }{3} \right)$
The value of $\cos \dfrac{\pi }{3}$ is $\dfrac{1}{2}$.
So the value $\cos \dfrac{2\pi }{3}=-\dfrac{1}{2}$ .
Since the result is the same as that of the question it means the answer is correct.