64 CHAPTER 2. FUNCTIONS

Therefore, for all n, the area of S (θ) is trapped between the two numbers,

r2

2θ

sin(θ/n)(θ/n)

,r2

2θ

tan(θ/n)(θ/n)

.

Now let ε > 0 be given, a small positive number less than 1, and let n be large enough that

1 ≥ sin(θ/n)(θ/n)

≥ 11+ ε

and1

1+ ε≤ tan(θ/n)

(θ/n)≤ 1

1− ε.

Therefore,r2

2θ

(1

1+ ε

)≤ Area of S (θ) ≤

(1

1− ε

)r2

2θ .

Since ε is an arbitrary small positive number, it follows the area of the sector equals r2

2 θ asclaimed. (Why?)

2.4 Exercises1. Find cosθ and sinθ using only knowledge of angles in the first quadrant for θ ∈{ 2π

3 , 3π

4 , 5π

6 ,π, 7π

6 , 5π

4 , 4π

3 , 3π

2 , 5π

3 , 7π

4 , 11π

6 ,2π}.

2. Prove cos2 θ =1+ cos2θ

2and sin2

θ =1− cos2θ

2.

3. π/12 = π/3−π/4. Therefore, from Problem 2,

cos(π/12) =

√1+(√

3/2)

2.

On the other hand,

cos(π/12) = cos(π/3−π/4) = cosπ/3cosπ/4+ sinπ/3sinπ/4

and so cos(π/12) =√

2/4+√

6/4. Is there a problem here? Please explain.

4. Prove 1+ tan2 θ = sec2 θ and 1+ cot2 θ = csc2 θ .

5. Prove that sinxcosy =12(sin(x+ y)+ sin(x− y)) .

6. Prove that sinxsiny =12(cos(x− y)− cos(x+ y)) .

7. Prove that cosxcosy =12(cos(x+ y)+ cos(x− y)) .

8. Using Problem 5, find an identity for sinx− siny.

9. Suppose sinx = a where 0 < a < 1. Find all possible values for