Consider an arc A, of a circle of radius r which subtends an angle θ. The circular sector
determined by A is obtained by joining the ends of the arc A, to the center of the
circle.

PICT

The sector, S

(θ)

denotes the points which lie between the arc A and the two lines just
mentioned. The angle between the two lines is called the central angle of the sector. The
problem is to define the area of this shape. First a fundamental inequality must be
obtained.

Lemma 2.3.12Let 1 > ε > 0 be given. Then whenever the positive number α, issmall enough,

α
1 ≤ sinα-≤ 1 + ε (2.14)

(2.14)

and

--α--
1 +ε ≥ tanα ≥ 1 − ε (2.15)

(2.15)

Proof: This follows from Corollary 2.3.10 on Page 149. In this corollary, l

(A)

= α
and so

1− cosα + sinα ≥ α ≥ sin α.

Therefore, dividing by sinα,

1 − cos α α
--sin-α--+ 1 ≥ sin-α ≥ 1. (2.16)

(2.16)

Now using the properties of the trig functions,

1−-cosα-
sinα

=

--1-−-cos2α----
sin α(1+ cosα)

=

sin2 α
sin-α(1+-cosα)-

=

sinα
1+-cosα-

.

From the definition of the sin and cos, whenever α is small enough,

--sinα---< ε
1+ cosα

and so 2.16 implies that for such α, 2.14 holds. To obtain 2.15, let α be small enough that
2.14 holds and multiply by cosα. Then for such α,

α
cosα ≤ -----≤ (1+ ε)cosα
tanα

Taking α smaller if necessary, and noting that for all α small enough, cosα is very close
to 1, yields 2.15. ■

This lemma is very important in another context.

Theorem 2.3.13Let S

(θ)

denote the sector of a circle of radius r havingcentral angle θ. Then the area of S

(θ)

equals

r2
2-

θ.

Proof:Let the angle which A subtends be denoted by θ and divide this sector into n
equal sectors each of which has a central angle equal to θ∕n. The following is a picture of
one of these.

PICT

In the picture, there is a circular sector, S

(θ∕n)

and inside this circular sector is a
triangle while outside the circular sector is another triangle. Thus any reasonable
definition of area would require

r2 r2
2 sin(θ∕n) ≤ area of S (θ∕n) ≤ 2 tan(θ∕n).

It follows the area of the whole sector having central angle θ must satisfy the following
inequality.

nr2 nr2
---sin(θ∕n) ≤ area of S (θ) ≤---tan(θ∕n).
2 2