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PROOF 32

Prove.

Trigonometric ratios are determined by the angle alone

On right triangles.

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

For an acute angle θ and a right triangle T having θ as one acute angle, s(θ, T) is the length of the side opposite θ in T divided by the length of the hypotenuse.

Proof.

(1)A right triangle having θ as one acute angle exists.A3
(2)In that triangle, the side opposite the right angle is called the hypotenuse, and the side opposite θ the height.convention
(3)The length of the hypotenuse is nonzero, so the division is defined.A4
(4)Set s(θ, T) = height ÷ hypotenuse. This value still depends on T.(2), (3), convention

∎

Lemma L1

Two right triangles having the same acute angle θ are similar.

Proof.

(1)Let T = △ABC have a right angle at C with ∠A = θ, and T′ = △A′B′C′ a right angle at C′ with ∠A′ = θ.assumption
(2)∠C = ∠C′ = 90°(1)
(3)∠A = ∠A′ = θ(1)
(4)Two pairs of angles are equal, so the triangles are similar under the correspondence A↔A′, B↔B′, C↔C′.(2), (3), A1

∎

Lemma L2

If △ABC and △A′B′C′ are similar under A↔A′, B↔B′, C↔C′, then BC/AB = B′C′/A′B′

Proof.

(1)Set a = BC, c = AB, a′ = B′C′, c′ = A′B′.convention
(2)a, c, a′, c′ are all greater than 0.A4
(3)BC and B′C′, AB and A′B′ are corresponding sides, so a/a′ = c/c′assumption, A2, (2)
(4)a c′ = a′ c(3), A5, (2)
(5)Dividing both sides by c c′ gives (a c′)/(c c′) = (a′ c)/(c c′)(4), (2), E
(6)c c′ = c′ c, so a/c = a′/c′(5), A7, A6

∎

Theorem T1

For any two right triangles T, T′ having the same acute angle θ, s(θ, T) = s(θ, T′)

Proof.

(1)Setting T = △ABC, T′ = △A′B′C′ as in L1, the two triangles are similar.L1
(2)∠C is the right angle, so the hypotenuse is AB; ∠A = θ, so the side opposite θ is BC.(1), D1
(3)In T′ likewise, the hypotenuse is A′B′ and the side opposite θ is B′C′.(1), D1
(4)BC/AB = B′C′/A′B′(1), L2
(5)s(θ, T) = BC/AB, s(θ, T′) = B′C′/A′B′(2), (3), D1
(6)s(θ, T) = s(θ, T′)(4), (5)

∎

Corollary C

sin θ is defined as a function of the angle θ alone.

Proof.

(1)A right triangle T with angle θ exists.A3
(2)The value of s(θ, T) is independent of the choice of T.T1
(3)Therefore setting sin θ = s(θ, T) assigns a single value to each angle θ.(1), (2), D1
(4)adjacent ÷ hypotenuse and opposite ÷ adjacent are likewise independent of T by the same argument, and are defined as cos θ and tan θ respectively.L1, L2, A3, A4

∎

Remark

Can A1 be removed from this proof?

Separate the line that passes from angles to similarity from the line that passes from similarity to ratios. They do not use the same grounds.