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You can show that by constructing a triangle.
Take two points, O(0, 0) and A(1, 0), and let B be the point on the unit circle such that the angle between the line segments OA and OB is radians.
Since both A and B lie on the circle, the line segments OA and OB both have length 1 (same as the circle's radius). We finish constructing the triangle by connect A and B.
Since OB and OA have the same length, triangle OAB is isosceles, but more than that, it's also equilateral. Why? Because the interior angles of any triangle always add to radians. We know one of the angles is
radians, which leaves a contribution of
radians between the remaining angles A and B. Angles A and B must be congruent (because OAB is isosceles), which means they also have measure
radians.
Next, draw an altitude of the triangle through point B, and label the point where it meets the "base" OA, C. Since OAB is equilateral, the altitude BC is also a perpendicular bisector. That means OC has length , and by definition of
we have
