Question

How do you find the value of ${\cos ^{ - 1}}( - \dfrac{{\sqrt 3 }}{2})$ ?

Answer 1

Trigonometry ratios are used to find the relation between two sides and an angle of a right-angled triangle as they are given as the ratio of two sides of the right angles triangle. In the given question, we have to find the value of ${\cos ^{ - 1}}( - \dfrac{{\sqrt 3 }}{2})$ , ${\cos ^{ - 1}}A$ means the cosine inverse of the value whose cosine is $A$ . Thus, we have to find the cosine inverse of the angle whose cosine is $\dfrac{3}{4}$ , we know that $A = \cos \theta = - \dfrac{{\sqrt 3 }}{2}$ , using trigonometric identities and formulas, we can find out the correct answer.

Complete step-by-step solution:
We know that cosine function is negative in the second and the third quadrant, and we also know that $\cos \dfrac{\pi }{6} = \dfrac{{\sqrt 3 }}{2}$, so –
$- \dfrac{{\sqrt 3 }}{2} = \cos (\pi - \dfrac{\pi }{3}),\, - \dfrac{{\sqrt 3 }}{2} = \cos (\pi + \dfrac{\pi }{3})$
Range of ${\cos ^{ - 1}}\theta$ is $[0,\pi ]$ , so $\theta$ cannot be smaller than 0 or greater than $\pi$ , thus $\cos (\pi + \dfrac{\pi }{3})$ is rejected.
So, $\cos \dfrac{{5\pi }}{6} = \dfrac{{ - \sqrt 3 }}{2}$
We know that
${\cos ^{ - 1}}(\cos \theta ) = \theta \\\ \Rightarrow {\cos ^{ - 1}}(\cos \dfrac{{5\pi }}{6}) = \dfrac{{5\pi }}{6} \\\ \Rightarrow {\cos ^{ - 1}}( - \dfrac{{\sqrt 3 }}{2}) = \dfrac{{5\pi }}{6} \\\$

Hence, the value of ${\cos ^{ - 1}}( - \dfrac{{\sqrt 3 }}{2})$ is $\dfrac{{5\pi }}{6}$ or $150^\circ$ .

Note: Before solving these types of questions, we should ensure that we have to find the general solution or the principal solution. In this question, we find the answer that lies between the principal branch that is $[0,\pi ]$ , so we ignore the general solution, the given question can have infinite general solutions. There are four quadrants in a graph and different trigonometric functions have different signs in different quadrants. The sign of cosine function is negative only in the second and third quadrant, an angle lying in the second quadrant is smaller than $\pi$ and greater than $\dfrac{\pi }{2}$ , and an angle lying in the third quadrant is smaller than $\dfrac{{3\pi }}{2}$and greater than $\pi$ that’s why we subtract the angle from $\pi$ and add the angle to $\pi$.