Question

How do you evaluate \[\arccos \left( {\dfrac{{\sqrt 3 }}{2}} \right)\] without using a calculator?

Answer 1

Hint: To solve this type question we will use the inverse trigonometric functions. We will first know the cos of which angle has the value so given in the question and the we will assign that value in the given function to find the evaluation.

Complete step-by-step answer:
We are asked to find the \[\arccos \left( {\dfrac{{\sqrt 3 }}{2}} \right)\]
We know that,
\[\arccos \left( {\dfrac{{\sqrt 3 }}{2}} \right) = {\cos ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right)\]
But \[\cos \dfrac{\pi }{6} = \dfrac{{\sqrt 3 }}{2}\]
So we will put this value in the given evaluation above,
\[ = co{s^{ - 1}}\left( {\cos \dfrac{\pi }{6}} \right)\]
We know that,
\[co{s^{ - 1}}\left( {\cos \theta } \right) = \theta \]
So we can write,
\[co{s^{ - 1}}\left( {\cos \dfrac{\pi }{6}} \right) = \dfrac{\pi }{6}\]
So, \[\arccos \left( {\dfrac{{\sqrt 3 }}{2}} \right) = \dfrac{\pi }{6}\]
This is the correct answer.
So, the correct answer is “$\dfrac{\pi }{6}$”.

Note: Note that we need not to make any heavy calculations here. We just need to assign the value of the cos function here in the bracket. Also note that arc stands for inverse function only. We have inverse function for remaining trigonometric functions also like for sin , tan, sec, cosec also.
Sometimes the angle is in degrees but sometimes it is in radians also. But the value wont change only the type of writing the angle changes.