Definition D1
a − b means a + (−b), a² means aa, and 2 means 1 + 1.
Proof.
(1)Subtraction is written as adding the additive inverse.convention
(2)The square and 2 are likewise abbreviations with these meanings.convention
∎
Lemma L1
For any real s, t, d(P(s), P(t))² = 2 − 2(cos s cos t + sin s sin t)
Proof.
(1)P(s) = (cos s, sin s), P(t) = (cos t, sin t)A4
(2)d(P(s), P(t))² = (cos s − cos t)² + (sin s − sin t)²(1), A1
(3)(cos s − cos t)² = cos²s − 2 cos s cos t + cos²tA11, D1
(4)(sin s − sin t)² = sin²s − 2 sin s sin t + sin²tA11, D1
(5)d² = (cos²s + sin²s) + (cos²t + sin²t) + ((−2 cos s cos t) + (−2 sin s sin t))(2), (3), (4), A13, A14, D1
(6)cos²s + sin²s = 1 and cos²t + sin²t = 1A5, A13
(7)1 + 1 = 2D1
(8)(−2 cos s cos t) + (−2 sin s sin t) = −(2(cos s cos t + sin s sin t))A17, A18, A12
(9)d(P(s), P(t))² = 2 − 2(cos s cos t + sin s sin t)(5), (6), (7), (8), D1
∎
Lemma L2
For any real α, β, d(P(α), P(β)) = d(P(α − β), P(0))
Proof.
(1)α + (−β) = α − β and β + (−β) = 0D1, A19
(2)Rotation by −β carries P(α) to P(α − β) and P(β) to P(0).(1), A3
(3)Rotation does not change the distance between two points.A2
(4)d(P(α), P(β)) = d(P(α − β), P(0))(2), (3)
∎
Theorem T1
For all real α, β, cos(α − β) = cos α cos β + sin α sin β
Proof.
(1)Setting s = α, t = β gives d(P(α), P(β))² = 2 − 2(cos α cos β + sin α sin β)L1
(2)Setting s = α − β, t = 0 gives d(P(α−β), P(0))² = 2 − 2(cos(α−β) cos 0 + sin(α−β) sin 0)L1
(3)cos 0 = 1, sin 0 = 0A6
(4)cos(α−β) × 1 = cos(α−β), sin(α−β) × 0 = 0, x + 0 = xA15, A16
(5)d(P(α−β), P(0))² = 2 − 2 cos(α−β)(2), (3), (4)
(6)The two distances are equal, so their squares are equal.L2, E
(7)2 + (−2(cos α cos β + sin α sin β)) = 2 + (−2 cos(α−β))(1), (5), (6), D1
(8)−2(cos α cos β + sin α sin β) = −2 cos(α−β)(7), A20
(9)−2 ≠ 0, so cos(α − β) = cos α cos β + sin α sin β(8), A22, A21
∎
Theorem T2
For all real α, β, cos(α + β) = cos α cos β − sin α sin β
Proof.
(1)−(−β) = β, so α + β = α − (−β)A18, D1
(2)cos(α − (−β)) = cos α cos(−β) + sin α sin(−β)T1
(3)cos(−β) = cos β, sin(−β) = −sin βA7, A8
(4)sin α(−sin β) = −(sin α sin β)A17
(5)cos(α + β) = cos α cos β − sin α sin β(1), (2), (3), (4), D1
∎
Theorem T3
For all real α, β, sin(α + β) = sin α cos β + cos α sin β
Proof.
(1)sin(α + β) = cos(90° − (α + β))A9
(2)90° − (α + β) = (90° − α) − βA13, A14, A18, D1
(3)cos((90° − α) − β) = cos(90° − α) cos β + sin(90° − α) sin βT1
(4)cos(90° − α) = sin α, sin(90° − α) = cos αA9, A10
(5)sin(α + β) = sin α cos β + cos α sin β(1), (2), (3), (4)
∎
Corollary C
sin 2α = 2 sin α cos α and cos 2α = cos²α − sin²α
Proof.
(1)Setting β = α in T3 gives sin(α + α) = sin α cos α + cos α sin αT3
(2)cos α sin α = sin α cos αA13
(3)x + x = x×1 + x×1 = x(1 + 1) = x × 2 = 2xA15, A12, A13, D1
(4)α + α = 2α, so sin 2α = 2 sin α cos α(1), (2), (3), D1
(5)Setting β = α gives cos 2α = cos α cos α − sin α sin α = cos²α − sin²αT2, (3), D1
∎