How do you evaluate \[\arcsin 1\]?

Hint: In the above question we are given an expression that is \[\arcsin 1\] and we are asked to evaluate it. By seeing the given expression that is \[\arcsin 1\] we can say that it is an inverse trigonometric function which are the inverse functions of the trigonometric functions like the sine, cosine, tangent, cotangent, secant and cosecant and are used to obtain an angle from the given trigonometric ratios that belong to a certain angle. In this question we firstly need to rewrite the given expression \[\arcsin 1\] in a simpler way then evaluate for the angle value. Let us see the method of evaluation of the given expression that is \[\arcsin 1\] in the complete step by step solution.

Complete Step By Step Solution :
In the question we are given an inverse trigonometric expression (inverse trigonometric expressions are the functions which are the inverse functions of the trigonometric functions like the sine, cosine, tangent, cotangent, secant and cosecant and are used to obtain an angle from the given trigonometric ratios that belong to a certain angle) that is \[\arcsin 1\] and we are asked the method to evaluate that is we need first write it in a simpler form that is –
\[
  \arcsin 1 \\
   = {\sin ^{ - 1}}1 = \theta \\
 \]
Also the range of the \[\arcsin \theta \]is \[\left[ {\dfrac{{ - \pi }}{2}, + \dfrac{\pi }{2}} \right]\]
Now further solving we get –
\[
  {\sin ^{ - 1}}1 = \theta \\
  \sin \theta = 1 \\
 \]
Here we supposed the \[{\sin ^{ - 1}}1\] is equal to \[\theta \]
That is for \[\theta \in \left[ {\dfrac{{ - \pi }}{2}, + \dfrac{\pi }{2}} \right]\] with a unit circle, the ratio of the opposite side(y-coordinate) to the hypotenuse is must be-\[ = 1\]
The only angle meeting this requirement is \[\theta = {90^\circ }or\dfrac{\pi }{2}\]

So. the solution of the given inverse trigonometric function \[\arcsin 1\] is \[\theta = {90^\circ }or\dfrac{\pi }{2}\]

Note :
While solving such kinds of questions one should know about the inverse trigonometric ranges of the different trigonometric functions and their values at different angles as it is the main key in solving such kinds of questions easily and correctly.