mr_e_man wrote:4*2arcsin(φ⁻¹) + 2arcsin(φ⁻¹/√2) + 2arcsin(φ⁻⁷/√2) = 360°
Dividing by 2 and subtracting:
arcsin(φ⁻¹/√2) + arcsin(φ⁻⁷/√2) ≟ 180° - 4arcsin(φ⁻¹)
The arcsine of any positive number is between 0° and 90°, so the left side of this equation is between 0° and 180°.
Also, since φ⁻¹ < 1/√2, it follows that arcsin(φ⁻¹) < 45°, so the right side is also between 0° and 180°.
The cosine function is invertible on this interval, so it can be applied without changing the truth of the equation:
cos(arcsin(φ⁻¹/√2) + arcsin(φ⁻⁷/√2)) ≟ cos(180° - 4arcsin(φ⁻¹))
Here we can use the addition law, cos(x+y)=cos(x)cos(y)-sin(x)sin(y), along with cos(arcsin(x))=√(1-x²), to expand the left side:
cos(arcsin(φ⁻¹/√2) + arcsin(φ⁻⁷/√2))
= √(1 - (φ⁻¹/√2)²) √(1 - (φ⁻⁷/√2)²) - (φ⁻¹/√2) (φ⁻⁷/√2)
= √(1 - φ⁻²/2) √(1 - φ⁻¹⁴/2) - φ⁻⁸/2
= √(1 - φ⁻²/2 - φ⁻¹⁴/2 + φ⁻¹⁶/4) - φ⁻⁸/2
And here's the right side:
cos(180° - 4arcsin(φ⁻¹))
= -cos(4arcsin(φ⁻¹))
= 1 - 2cos²(2arcsin(φ⁻¹))
= 1 - 2(1 - 2sin²(arcsin(φ⁻¹)))²
= 1 - 2(1 - 2φ⁻²)² = 1 - 2(2φ - 3)²
= 16φ - 25
So the equation becomes:
√(1 - φ⁻²/2 - φ⁻¹⁴/2 + φ⁻¹⁶/4) - φ⁻⁸/2 ≟ 16φ - 25
Multiplying by 2φ⁸ and adding 1:
√(4φ¹⁶ - 2φ¹⁴ - 2φ² + 1) ≟ 32φ⁹ - 50φ⁸ + 1
√(4(987φ+610) - 2(377φ+233) - 2(φ+1) + 1) ≟ 32(34φ+21) - 50(21φ+13) + 1
√(3192φ + 1973) ≟ 38φ + 23
Both sides are obviously positive, so squaring is invertible and thus doesn't change the truth of the equation:
3192φ + 1973 ≟ (38φ + 23)²
3192φ + 1973 ≟ (38² + 2*38*23)φ + (38² + 23²)
At last we see that it is true.
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