Question

Find the principal value of ${\cos }^{-1}\left({\displaystyle \frac{\sqrt{3}}{2}}\right)$.

Answer 1

Hint: The principal value of an inverse trigonometric function at a point x is the value of the inverse function at the point x, which lies in the range of the principal branch. The principal value of the cosine function is from $\left[0,\pi \right]$. From the table we will find the value of the function within the principal range, to get our final answer.

Complete step-by-step answer:

The values of the cosine function are-

Function	$0^o$	${30}^o$	${45}^o$	${60}^o$	${90}^o$
cos	1	$${\displaystyle \frac{\sqrt{3}}{2}}$$	$${\displaystyle \frac{1}{\sqrt{2}}}$$	$${\displaystyle \frac{1}{2}}$$	0

In the given question we need to find the principal value of ${\cos }^{-1}\left({\displaystyle \frac{\sqrt{3}}{2}}\right)$. We know that for the cosine function to be positive, the angle should be acute, that is less than ${90}^o$.
We know that $\cos {30}^o={\displaystyle \frac{\sqrt{3}}{2}}$.
${\cos }^{-1}\left({\displaystyle \frac{\sqrt{3}}{2}}\right)={30}^o$
We know that $\pi$ rad = ${180}^o$, so
${\cos }^{-1}\left({\displaystyle \frac{\sqrt{3}}{2}}\right)={\displaystyle \frac{\pi }{6}}$
This is the required value.

Note: In such types of questions, we need to strictly follow the range of the principal values that have been specified. The principal value of the cosine function in this case ranges from $0^o$ to ${90}^o$. This is because there can be infinite values of any inverse trigonometric functions.One common mistake is that students get confused between the sine and the cosine functions, and often write the values of sine function or vice versa.