Sherwood P. answered • 09/18/20

Certificated Math Teacher and Knowledgeable Precalculus Tutor

Problem statement: Find the exact value of arctan( sin (PI/2) ).

First, notice that the angle Pi/2 is in units of **radians**, not degrees. 2*Pi radians = 360 degrees.

**Pi/2 radians = 90 degrees.**

Second, review the definition of the functions f(u) = sin(u) and g(v) = arctan(v). I am using u and v as variable names so that they don't get confused with the coordinates x and y in the Cartesian coordinate system.

sin(u) = the length of the side of a right triangle opposite the angle u, where 1 is the length of the hypotenuse of that right triangle. For angles of u > 90 degrees but less than 270 degrees, sin(u) will be negative. For angles of u > 270 degrees but less than 360 degrees, u will be positive.

This is because the hypotenuse of the right triangle used to calculate sin(u) traces out a unit circle going counter-clockwise around the origin of the Cartesian coordinate system. The (x,y) points on this unit circle = (cos(u),sin(u)). The angle u is measured counter-clockwise starting at 0 for the positive x-axis.

When u = Pi/2, the point (x,y) on the unit circle = (0,1). So sin(Pi/2) = sin(90 degrees) = 1.

arctan(v) = the size of the angle g in a right triangle, where the ratio of the length of the side opposite angle g* (this length = the y coordinate of the point on a unit circle for angle g) *divided by the length of the side adjacent to angle g *(= the x coordinate of this point on the unit circle) *= v. Because there are two points on the unit circle that can have the same ratio y/x, the range of the function arctan(v) is limited to -Pi/2 < arctan(v) < Pi/2.

When v =1, this means y/x = 1 for (x,y) coordinates on the unit circle. This occurs at the point (sqrt(2)/2,sqrt(2)/2) . *Notice that (-**sqrt(2)/2,-sqrt(2)/2) is out of range.*

The angle g(1) = arctan(1) = **arctan( sin(Pi/2) ) = 45 degrees = Pi/4 **