2.3. CIRCULAR FUNCTIONS 63

and1+ ε ≥ α

tanα≥ 1− ε (2.15)

Proof: This follows from Corollary 2.3.9 on Page 59. In this corollary, l (A) = α andso

1− cosα + sinα ≥ α ≥ sinα.

Therefore, dividing by sinα,

1− cosα

sinα+1 ≥ α

sinα≥ 1. (2.16)

Now using the properties of the trig functions,

1− cosα

sinα=

1− cos2 α

sinα (1+ cosα)=

sin2α

sinα (1+ cosα)=

sinα

1+ cosα.

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

1+cosα< ε and so 2.16

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

cosα ≤ α

tanα≤ (1+ ε)cosα

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

This lemma is very important in another context.

Theorem 2.3.12 Let S (θ) denote the sector of a circle of radius r having central

angle θ . Then the area of S (θ) equalsr2

2θ .

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

r

S(θ/n)

In the picture, there is a circular sector, S (θ/n) and inside this circular sector is atriangle while outside the circular sector is another triangle. Thus any reasonable definitionof area would require

r2

2sin(θ/n)≤ area of S (θ/n)≤ r2

2tan(θ/n) .

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

nr2

2sin(θ/n)≤ area of S (θ)≤ nr2

2tan(θ/n) .