Series Question Ib 2004?

1. Anonymous Says

   I will write |XY| for the length of the line segment XY. From the definition of the trigonometric functions (look at the right triangle AOB_1) we have cos(theta) = |OB_1|/|OA|. Since the circle AOB has radius 1, |OA| = 1, we deduce cos(theta) = |OB_1|. Since arc A_1 B_1 is an arc of a circle with center O we deduce |OB_1| = |OA_1| and hence
   |OA_1| = cos(theta).
   From the definition of the trigonometric functions (look at the right triangle A_1OB_2) we have cos(theta) = |OB_2|/|OA_1|. Since |OA_1| = cos(theta), we deduce that cos(theta) = |OB_2|/cos(theta) and hence |OB_2| = (cos(theta))^2. Since arc A_2 B_2 is an arc of a circle with center O we deduce |OB_2| = |OA_2| and hence
   |OA_2| = (cos(theta))^2.
   Continuing on in this way, you find that
   |OA_n| = (cos(theta))^n for any positive integer n.
   By definition of the radian measure of an angle, the length s of an arc cut off in a circle of radius r by a wedge of angle theta are related by the equation s = r theta. Since arc AB is cut off in a circle of radius 1 by a wedge of angle theta we have (length of arc AB) = 1 theta. Since arc A_1 B_1 is cut off in a circle of radius |OA_1| = cos(theta) by a wedge of angle theta we have (length of arc A_1 B_1) = cos(theta) theta. Since arc A_2 B_2 is cut off in a circle of radius |OA_2| = (cos(theta))^2 by a wedge of angle theta we have (length of arc A_1 B_1) = (cos(theta))^2 theta. Continuing in this way you find that
   (length of arc A_n B_n) = (cos(theta))^n theta for any positive integer n,
   So your sum is
   theta + cos(theta) theta + (cos(theta))^2 theta + … + (cos(theta))^n theta + …
   This is a geometric series with first term a = theta and common ratio of terms r = cos(theta). Its sum, by the geometric series formula, is
   a/(1 – r) = theta/(1 – cos(theta)).

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