## Sunday, January 21, 2007

### Solving Van Roomen's Problem: Step Four

In today's blog, I will present some simple trigonometric identities that can be used to simplify Van Roomen's examples of x and A values. For those who want an introduction to Van Roomen's Problem, start here. In my next blog, I will apply these lemmas to Van Roomen's problem.

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) - 230 + 6√5 - 2√5 = (1/16)(36 + 4√5 + 2(√30 + 6√5)(5) - 230 + 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) + 230 + 6√5 ] =

= (1/16)[28 - 4√5 - 2(√30 + 6√5)(5) + 230 + 6√5] =

= (1/16)[28
- 4√5 - (30 + 6√5)(25 - 2)] =

=
(1/16)[28 - 4√5 - (√30 + 6√5)(√2√5 - 2)2] =

= (1/16)[28 - 4√5 - (30 + 6√5)(√20 - 85 + 4)] =

=
(1/16)[28 - 4√5 - (30 + 6√5)(√24 - 85)] =

=
(1/16)[28 - 4√5 - (√720 - 240√5 + 1445 - 240)] =

= (1/16)[28 - 4√5 - (√480 - 96√5] =

= (1/16)[28 - 4√5 - 4(30 - 65]

(4) 2sin(π/15) = √(1/16)[28 - 4√5 - 4(√30 -65] =

= (1/4)√[28 - 4√5 - 4(√30 -65] =

= (1/2)√[7 - √5 - √30 -65] =

QED

References
• Jean-Pierre Tignol, Galois' Theory of Algebraic Equations , World Scientific, 2001