Note 67 — PART SECOND. NOTES ON THE ILLUMINATION OF LIGHTHOUSES, WITH SHORT NOTICES OF THEIR EARLY HISTORY. (continued 2)
Reading edition
This expression is equivalent to that of M. Fresnel, but owing to a simplification in the fractional coefficients, it is not literally the same. I was led to it by the following steps, starting from the original equation sin ξ = m sin (2 ξ - θ)
sin ξ = m sin (2 ξ - θ)
= m{sin 2 ξ . cos θ - cos 2 ξ sin θ}
= m cos θ . sin 2 ξ - m sin θ . cos 2 ξ
= m cos θ . 2 sin ξ . cos ξ - m sin θ . {1 - 2 sin² ξ}
= 2 m cos θ . sin ξ . cos ξ - m sin θ + 2 m sin θ . sin² ξ.
Therefore, m sin θ + sin ξ - 2 m sin θ sin² ξ = 2 m cos θ . sin ξ . cos ξ.
Then:
m² sin² θ + 2 m sin θ . sin ξ - 4 m² sin² θ sin² ξ + sin² ξ - 4 m sin θ . sin³ ξ + 4 m² sin² θ . sin⁴ ξ = 4 m² cos² θ sin² ξ (1 - sin² ξ) = 4 m² . cos² θ . sin² ξ - 4 m² cos² θ sin⁴ ξ.
Hence we have:
m² sin² θ + 2 m sin θ . sin ξ + (1 - 4 m²) . sin² ξ - 4 m sin θ . sin³ ξ + 4 m² sin⁴ ξ = 0
Then dividing by 4 m² and arranging according to powers of ξ, we have as above:
sin⁴ ξ - 1 m sin θ . sin³ ξ + (1 4 m² - 1) . sin² ξ + 12 m . sin θ . sin ξ + ¹⁄₄ sin² θ = 0