Question

How do you evaluate $\arcsin \left( {\dfrac{{\sqrt 3 }}{2}} \right)$ ?

Answer 1

Hint : Given is a trigonometric function sine. This is one of the main trigonometric functions. There are different values for this function for different angles. The question we are given is asking for the inverse value of a sine function with the given value such that we have to find the angle that has value so given. So we will first assign the function with the respective value and then take the inverse of that.

Complete step-by-step answer :
Given that,
$\arcsin \left( {\dfrac{{\sqrt 3 }}{2}} \right)$ is the function given.
But we know that, $\sin \dfrac{\pi }{3} = \dfrac{{\sqrt 3 }}{2}$
Now taking the sine function on other side,
$\dfrac{\pi }{3} = \dfrac{1}{{\sin }}\left( {\dfrac{{\sqrt 3 }}{2}} \right)$
So simplifying this we get,
$\dfrac{\pi }{3} = {\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right)$
That is,
${\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right) = \dfrac{\pi }{3}$
Thus the value is $\dfrac{\pi }{3}$ or ${60^ \circ }$ .
This is our answer.
So, the correct answer is “ $\dfrac{\pi }{3}$ or ${60^ \circ }$ ”.

Note : Note that arc is nothing but the inverse functions so don’t get confused. So we have found the angle that is having the value of $\dfrac{{\sqrt 3 }}{2}$ . If this type of question appears in multiple choices and with options in degrees and radians both the both are correct answers. One answer is in degrees and the other is in radians. Such that $2\pi = {360^ \circ }$ .
Also note that,
$\cos ec\theta = \dfrac{1}{{\sin \theta }}$ but if $\theta = {60^ \circ }$ then $\cos ec{60^ \circ } = \dfrac{2}{{\sqrt 3 }}$ and $\sin {60^ \circ } = \dfrac{{\sqrt 3 }}{2}$ .
This is the difference between ${\sin ^{ - 1}}$ and $\cos ec\theta = \dfrac{1}{{\sin \theta }}$ .