Kenneth Kuttler

62 CHAPTER 2. FUNCTIONS

|P′1|+ ε > l (A1) , |P′

2|+ ε > l (A2) , and both P′1 and P′

2 determine the same sequence ofangles. Using Problem 36 on Page 68 about the base angles of an isosceles triangle, thetwo triangles are similar and so

|P′1|∣∣P′2

∣∣ = Rr

Thereforel (A2)<

∣∣P′2∣∣+ ε =

∣∣P′1∣∣+ ε ≤ r

Rl (A1)+ ε.

Since ε is arbitrary, this shows Rl (A2)≤ rl (A1) . But now reverse the argument and write

l (A1)<∣∣P′

1∣∣+ ε =

∣∣P′2∣∣+ ε ≤ R

rl (A2)+ ε

which implies, since ε is arbitrary that Rl (A2)≥ rl (A1) and this has proved the followingtheorem.

Theorem 2.3.10 Let θ be an angle which subtends two arcs, AR on a circle ofradius R and Ar on a circle of radius r. Then denoting by l (A) the length of a circular arcas described above, Rl (Ar) = rl (AR) .

Now, with this theorem, one can prove the fundamental inequality of Corollary 2.3.9.Proof: Situate the angle θ such that one side is on the positive x axis and extend the

other side till it intersects the unit circle at the point (cosθ ,sinθ) . Then denoting the re-sulting arc on the circle by A, it follows that for all P ∈ P (A) the inequality (1− cosθ)+sinθ ≥ |P| ≥ sinθ . It follows that (1− cosθ) + sinθ is an upper bound for all the |P|where P ∈ P (A) and so (1− cosθ)+ sinθ is at least as large as the sup or least upperbound of the |P| . This proves the top half of the inequality. The bottom half follows be-cause l (A) ≥ L where L is the length of the line segment joining (cosθ ,sinθ) and (1,0)due to the definition of l (A) . However, L ≥ sinθ because L is the length of the hypotenuseof a right triangle having sinθ as one of the sides.

2.3.3 The Area of a Circular SectorConsider an arc A, of a circle of radius r which subtends an angle θ . The circular sectordetermined by A is obtained by joining the ends of the arc A, to the center of the circle.

S(θ) A

The sector, S (θ) denotes the points which lie between the arc A and the two lines justmentioned. The angle between the two lines is called the central angle of the sector. Theproblem is to define the area of this shape. First a fundamental inequality must be obtained.

Lemma 2.3.11 Let 1 > ε > 0 be given. Then whenever the positive number α, is smallenough,

1 ≤ α

sinα≤ 1+ ε (2.14)

62 CHAPTER 2. FUNCTIONS|P}|+€ > 1(A1),|P3| +€ > 1 (Az), and both Pi and P} determine the same sequence ofangles. Using Problem 36 on Page 68 about the base angles of an isosceles triangle, thetwo triangles are similar and soTherefore , rI(A2) < |i] +e= R |Pi|+e< plate:Since € is arbitrary, this shows Ri (Az) < r/(A,). But now reverse the argument and writeR R1(A1) < |Pi|+e= - |P;|-+e < —1(A2) +€which implies, since € is arbitrary that R/ (Az) > rl (Aj) and this has proved the followingtheorem.Theorem 2.3.10 Let 6 be an angle which subtends two arcs, Ar on a circle ofradius R and A, on a circle of radius r. Then denoting by 1(A) the length of a circular arcas described above, RI(A,) = rl (Ar).Now, with this theorem, one can prove the fundamental inequality of Corollary 2.3.9.Proof: Situate the angle @ such that one side is on the positive x axis and extend theother side till it intersects the unit circle at the point (cos 0,sin@). Then denoting the re-sulting arc on the circle by A, it follows that for all P € # (A) the inequality (1 — cos @) +sin@ > |P| > sin@. It follows that (1 —cos@)+sin@ is an upper bound for all the |P|where P € Y(A) and so (1—cos@) +sin@ is at least as large as the sup or least upperbound of the |P|. This proves the top half of the inequality. The bottom half follows be-cause /(A) > L where L is the length of the line segment joining (cos 0,sin@) and (1,0)due to the definition of / (A). However, L > sin @ because L is the length of the hypotenuseof a right triangle having sin @ as one of the sides. Jj2.3.3 The Area of a Circular SectorConsider an arc A, of a circle of radius r which subtends an angle @. The circular sectordetermined by A is obtained by joining the ends of the arc A, to the center of the circle.The sector, S(@) denotes the points which lie between the arc A and the two lines justmentioned. The angle between the two lines is called the central angle of the sector. Theproblem is to define the area of this shape. First a fundamental inequality must be obtained.Lemma 2.3.11 Let 1 > € > 0 be given. Then whenever the positive number a, is smallenough,a1<-sind<1it+e (2.14)