The content in today's blog is taken from Jean-Pierre Tignol's Galois' Theory of Algebraic Equations.
Let me start by introducting the some basic ideas from trigonometry.
Lemma 1: 2 cos(α/2) = ± √2 + 2 cos α
Proof:
(1) cos(2x) = 2 cos2(x) - 1 [See Corollary 3.1, here]
(2) Let x = α/2, then we get:
cos(α) = 2cos2(α/2) - 1
(3) Rearranging terms gives us:
2cos2(α/2) = cos(α) + 1
(4) Multiplying both sides by 2 gives us:
4cos2(α/2) = 2cos(α) + 2
(5) Finally, taking the square root of both sides gives us:
2cos(α/2) = ±√2 + 2cos(α)
QED
Lemma 2: 2 sin(α/2) = ± √2 - 2 cos α
Proof:
(1) cos(2x) = 1 - 2 sin2(x) [See Corollary 3.1, here]
(2) Let x = α/2, then we get:
cos(α) = 1 - 2sin2(α/2)
(3) Rearranging terms gives us:
2sin2(α/2) = 1 - cos(α)
(4) Multiplying both sides by 2 gives us:
4sin2(α/2) = 2 - 2cos(α)
(5) Finally, taking the square root of both sides gives us:
2sin(α/2) = ±√2 - 2cos(α)
QED
Lemma 3: 2 cos(π/3) = 1
Proof:
(1) 2cos(π/3) = 2cos(60°) [See here for review of radians]
(2) 2cos(60°) = 2(1/2) = 1 [See Corollary 1.2, here]
QED
Lemma 4: 2 sin(π/3) = √3
Proof:
(1) 2sin(π/3) = 2sin(60°) [See here for review of radians]
(2) 2sin(60°) = 2*(√3/2) = √3 [See Corollary 1.2, here]
QED
Lemma 5: 2cos(2π/5) = (√5 - 1)/2
Proof:
(1) cos (2π/5) = (√5 - 1)/4 [See Corollary 1.1, here]
(2) So:
2cos(2π/5) = (√5 - 1)/2
QED
Lemma 6: 2sin(2π/5) = √(5 + √5)/2
Proof:
(1) sin (2π/5) = √5 + √5/8 [See Corollary 1.1, here]
(2) So:
2*sin (2π/5) = 2(1/2)√5 + √5/2 = √(5 + √5)/2
QED
Lemma 7: 2cos(π/15) = (1/4)(√30 + 6√5 + √5 - 1)
Proof:
(1) 2π/5 - π/3 = 6π/15 - 5π/15 = π/15
(2) 2cos(π/15) = 2cos(2π/5 - π/3) = 2cos(2π/5)cos(π/3) + 2sin(2π/5)sin(π/3) [See Theorem 2, here]
2cos(2π/5) = (√5 - 1)/2 [See Lemma 5 above]
cos(π/3) = 1/2 [See Lemma 3 above]
2sin(2π/5) = √(5 + √5)/2 [See Lemma 6 above]
sin(π/3) = √3/2 [See Lemma 4 above]
(4) 2cos(π/15) = [ (√5 - 1)/2 ][1/2] + [√(5 + √5)/2][√3/2] =
= [ (√5 - 1)/4] +(1/2) [√(15 + 3√5)/2] = [ (√5 - 1)/4] + (1/4)[√(30 + 6√5)]
= (1/4)[ √5 - 1 + √(30 + 6√5)]
QED
Corollary 7.1: 2sin(π/15) = (1/2)√7 - √5 - √(30 - 6√5)
Proof:
(1) 2cos(π/15) = (1/4)(√30 + 6√5 + √5 - 1) [See Lemma 7 above]
(2) 4cos2(π/15) = (1/16)(30 + 6√5 + 5 + 1 + 2(√30 + 6√5)(√5) - 2√30 + 6√5 - 2√5 = (1/16)(36 + 4√5 + 2(√30 + 6√5)(√5) - 2√30 + 6√5)
(3) Since sin2(x) + cos2(x) = 1 [See Corollary 2, here], we have:
4sin2(π/15) = 4 - cos2(π/15) =
= (1/16)[64 - 36 - 4√5 - 2(√30 + 6√5)(√5) + 2√30 + 6√5 ] =
= (1/16)[28 - 4√5 - 2(√30 + 6√5)(√5) + 2√30 + 6√5] =
= (1/16)[28 - 4√5 - (√30 + 6√5)(2√5 - 2)] =
= (1/16)[28 - 4√5 - (√30 + 6√5)(√2√5 - 2)2] =
= (1/16)[28 - 4√5 - (√30 + 6√5)(√20 - 8√5 + 4)] =
= (1/16)[28 - 4√5 - (√30 + 6√5)(√24 - 8√5)] =
= (1/16)[28 - 4√5 - (√720 - 240√5 + 144√5 - 240)] =
= (1/16)[28 - 4√5 - (√480 - 96√5] =
= (1/16)[28 - 4√5 - 4(√30 - 6√5]
(4) 2sin(π/15) = √(1/16)[28 - 4√5 - 4(√30 -6√5] =
= (1/4)√[28 - 4√5 - 4(√30 -6√5] =
= (1/2)√[7 - √5 - √30 -6√5] =
QED
References
- Jean-Pierre Tignol, Galois' Theory of Algebraic Equations, World Scientific, 2001
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