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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