Definition D1
For an acute angle θ, in a right triangle having θ, sin θ is the side opposite θ divided by the hypotenuse, and cos θ is the side adjacent to θ divided by the hypotenuse.
Proof.
(1)A right triangle with θ exists, so the two values can be taken.A3
(2)The two values are the same in whichever triangle they are measured.A2
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Definition D2
For any real θ, let P(θ) be the point where the ray making angle θ counterclockwise from the positive x-axis meets the unit circle, and for its coordinates (x, y) define cos θ = x, sin θ = y.
Proof.
(1)For each angle θ there is exactly one such point.A5
(2)P(θ) lies on the unit circle, so OP(θ) = 1(1), A4
(3)This definition gives values for every real θ, beyond the acute angles. That it does not conflict with D1 on acute angles is verified in L2.(1), convention
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Lemma L1
A point (x, y) on the unit circle satisfies x² + y² = 1.
Proof.
(1)If P = (x, y) is a point on the unit circle, then OP = 1A4
(2)Suppose x ≠ 0 and y ≠ 0. Let H = (x, 0).assumption
(3)OH is a horizontal segment with OH = |x|; HP is a vertical segment with HP = |y|(2), A7
(4)The two segments are perpendicular, so triangle OHP has a right angle at H and its hypotenuse is OP.(3), A8
(5)OP² = OH² + HP²(4), A1
(6)1² = |x|² + |y|², so 1 = x² + y²(1), (3), (5), A9, A10
(7)Suppose y = 0. Then OP is a horizontal segment, so |x| = 1(1), A7
(8)x² = |x|² = 1 and y² = 0, so x² + y² = 1 + 0 = 1(7), A9, A10, A11
(9)The case x = 0 holds by the same argument with a vertical segment.(1), A7, A9, A10, A11
(10)In every case, x² + y² = 1(6), (8), (9)
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Lemma L2
If 0° < θ < 90°, then the cos θ, sin θ given by D2 equal the values given by D1.
Proof.
(1)Setting P(θ) = (x, y) gives x > 0, y > 0D2, A6
(2)Setting H = (x, 0) gives OH = |x| = x, HP = |y| = y(1), A7, A9
(3)The two segments are perpendicular, so triangle OHP has a right angle at H and its hypotenuse is OP = 1.(2), A8, D2
(4)H lies on the positive x-axis, so ∠HOP = θ(1), D2
(5)In triangle OHP, the side opposite θ is HP and the adjacent side is OH.(3), (4)
(6)By D1, sin θ = HP ÷ OP = y ÷ 1 = y, cos θ = OH ÷ OP = x ÷ 1 = x(2), (3), (5), D1, A10
(7)D2 also gives sin θ = y, cos θ = x, so the two definitions agree.(6), D2
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Theorem T1
For every real θ, sin²θ + cos²θ = 1
Proof.
(1)P(θ) = (cos θ, sin θ) is a point on the unit circle.D2
(2)cos²θ + sin²θ = 1(1), L1
(3)sin²θ + cos²θ = 1(2), A12
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Corollary C
In a right triangle with acute angle θ as well, (opposite ÷ hypotenuse)² + (adjacent ÷ hypotenuse)² = 1
Proof.
(1)For an acute angle θ, the values of D1 and D2 are equal.L2
(2)For every θ, sin²θ + cos²θ = 1T1
(3)Therefore it also holds for the two ratios given by D1.(1), (2), D1
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