Let's prove that lim |sin(t)| = O . t>0 Let D = [1, 0] and B be a point on the unit circle with centre A = [0, 0] . Let C be the projection of B onto the line AD , and let t be the angle ABAD in radians, as shown below. C3 1 = 0.88 1.5 2 2.5 - The length of the straight line segment BC can be expressed in terms oft as abs(sin(t)) o a; :5] (we require lengths to be positive or zero. so use an absolute value). - The length of the circular arc BD can be expressed in term of t as sin(t) o a . [:1 By comparing these lengths, we get the inequality 0 S lsin(t)| S ltl