comparison of and near
Theorem 1.
Let , where is an angle measured in radians. Then .
Proof.
Let , , and . Note that the circle passes through and and that the shortest arc along this circle from to has length . Note also that the line segments and are radii of the circle and therefore must each have length 1.
Draw the line segment . Since this does not correspond to the arc, its length .
Drop the perpendicular from to . Let be the point of intersection
. Note that and .
Since , and . Thus, and lies strictly in between and . Therefore, .
By the Pythagorean theorem, . Thus, . Therefore, .∎
The analogous result for slightly below 0 is:
Corollary 1.
Let , where is an angle measured in radians. Then .
Proof.
Since , the previous theorem yields . Since is an odd function, . It follows that .∎