PROOF 128

Prove.

The period of the trigonometric functions comes from one full turn

On the unit circle.

A way to read this proof

First separate what is assumed from what is to be proved. At each step, ask which definition or earlier result permits the move. If it is unclear, return to that premise rather than skipping to the conclusion.

Definition D1

The circle of radius 1 centered at the origin is called the unit circle. Let P(θ) be the point where the ray making angle θ counterclockwise from the positive x-axis meets the unit circle, and for P(θ) = (x, y) define cos θ = x, sin θ = y.

Proof.

(1)For each real θ there is exactly one such point, so cos and sin are functions.A1

(2)−(x, y) means (−x, −y).convention

∎

Definition D2

For θ with cos θ ≠ 0, define tan θ = sin θ ÷ cos θ.

Proof.

At θ with cos θ = 0, tan θ is not defined.convention

∎

Definition D3

If T > 0 and f(θ + T) = f(θ) for every θ in the domain, then T is called a period of f. The smallest period is called the fundamental period.

Proof.

A period is the interval at which a function repeats the same values.convention

∎

Theorem T1

2π is a period of cos and sin. That is, for every θ, cos(θ + 2π) = cos θ and sin(θ + 2π) = sin θ.

Proof.

(1)0 < π < 2π, so 2π > 0.A8

(2)P(θ + 2π) = P(θ)A2

(3)P(θ + 2π) = (cos(θ+2π), sin(θ+2π)) and P(θ) = (cos θ, sin θ)D1

(4)The two ordered pairs are equal, so cos(θ+2π) = cos θ, sin(θ+2π) = sin θ(2), (3), A5

(5)This holds for every θ, so 2π is a period of both functions.(1), (4), D3

∎

Theorem T2

No T with 0 < T < 2π can be a common period of cos and sin.

Proof.

(1)Suppose 0 < T < 2π and cos(θ+T) = cos θ, sin(θ+T) = sin θ for every θ.assumption, D3

(2)The two coordinates agree, so P(θ + T) = P(θ)(1), D1, A5

(3)Substituting θ = 0 gives P(T) = P(0)(2)

(4)0 ≤ 0 < T < 2π, so P(0) ≠ P(T)(1), A4

(5)(3) and (4) cannot both hold, so no such T exists.(3), (4)

∎

Corollary C1

2π is the fundamental period of cos and sin.

Proof.

(1)2π is a period.T1

(2)There is no common period T with 0 < T < 2π.T2

(3)Therefore 2π is the smallest common period, that is, the fundamental period.(1), (2), D3

∎

Corollary C2

If cos θ ≠ 0, then tan(θ + π) = tan θ. That is, π is a period of tan.

Proof.

(1)P(θ + π) = −P(θ) = (−cos θ, −sin θ)A3, D1

(2)The two ordered pairs are equal, so cos(θ+π) = −cos θ, sin(θ+π) = −sin θ(1), A5

(3)cos θ ≠ 0, so −cos θ ≠ 0, and tan(θ+π) is defined.(2), A7, D2

(4)tan(θ+π) = (−sin θ) ÷ (−cos θ)(2), (3), D2

(5)(−sin θ) ÷ (−cos θ) = sin θ ÷ cos θ = tan θ(4), A6, D2

(6)0 < π, so π is a period of tan.(5), A8, D3

(7)At the antipodal point both coordinates flip sign together, so their ratio does not change.(2), (5)

∎

Remark

Can A2 be removed from this proof?

Separate the lines that determine values from the lines that repeat values. What happens within one turn and what happens beyond one turn rest on different axioms.