Question

Determine the principal value of ${\cos ^{ - 1}}\left( { - \dfrac{1}{2}} \right)$.

Answer 1

In order to find the solution of a trigonometric equation, we start by taking the inverse trigonometric function like inverse sin, inverse cosine, inverse tangent on both sides of the equation and then set up reference angles to find the rest of the answers.

Complete step by step answer:
For ${\sin ^{ - 1}}$ function, the principal value branch is $\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$.For the ${\cos ^{ - 1}}$ function, the principal value branch is $\left[ {0,\pi } \right]$. For ${\tan ^{ - 1}}$ function, the principal value branch is $\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$. The principal value of ${\cos ^{ - 1}}\left( { - \dfrac{1}{2}} \right)$ will lie in the principal branch of cosine. Also, we must remember values of trigonometric functions for some standard angles in order to solve the problem.
According to definition of inverse ratio,
If$\cos x = - \dfrac{1}{2}$,
Then, ${\cos ^{ - 1}}\left( { - \dfrac{1}{2}} \right) = x$ where the value of x lies in the range $\left[ {0,\pi } \right]$.
Now, we know that the cosine function is positive in the first and fourth quadrants and negative in the second and in the third quadrant.
So, the angle $x = {\cos ^{ - 1}}\left( { - \dfrac{1}{2}} \right)$ must lie either in the second quadrant or in the third quadrant.
We know that the value of $\cos \left( {\dfrac{\pi }{3}} \right)$ is $\left( {\dfrac{1}{2}} \right)$. Also, $\cos \left( {\pi - \theta } \right) = - \cos \left( \theta \right)$.
So, $\cos \left( {\pi - \dfrac{\pi }{3}} \right) = - \cos \left( {\dfrac{\pi }{3}} \right)$ .
Therefore, $\cos \left( {\dfrac{{2\pi }}{3}} \right) = \left( { - \dfrac{1}{2}} \right)$.

Hence, the value of ${\cos ^{ - 1}}\left( { - \dfrac{1}{2}} \right)$ is $\left( {\dfrac{{2\pi }}{3}} \right)$.

Note: The basic inverse trigonometric functions are used to find the missing angles in right triangles. While the regular trigonometric functions are used to determine the missing sides of the right-angled triangles, using the following formulae:
$\sin \theta = \left( {\dfrac{{{\text{Opposite Side}}}}{{{\text{Hypotenuse}}}}} \right)$
$\Rightarrow \cos \theta = \left( {\dfrac{{{\text{Adjacent Side}}}}{{{\text{Hypotenuse}}}}} \right)$
$\Rightarrow \tan \theta = \left( {\dfrac{{{\text{Opposite Side}}}}{{{\text{Adjacent Side}}}}} \right)$
Besides the trigonometric functions and inverse trigonometric functions, we also have some rules related to trigonometry such as the sine rule and cosine rule. We must remember the principal value branches of the inverse trigonometric functions to tackle such problems. There is only one principal solution of inverse trigonometric function. Any value of inverse function other than the principal value is known as the general solution of inverse function.