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
The solution of this equation, which is of the fourth degree, is somewhat tedious; but as the root, which will satisfy the optical conditions of the question, must be the sine of an angle, and necessarily lies between zero and unity; and as the protraction, if conducted with due care in the manner already described, affords the means of at once assuming a probable value of ξ not very distant from the truth, the labour of the calculation, in this particular case, is not quite so great as might be expected. But notwithstanding all the abridgments of which the particular case admits, a considerable amount of labour is required, and a corresponding risk of error incurred, in merely introducing the numerical values into the equation preparatory to its solution; and any other method requiring less arithmetical operation, is, of course, greatly to be preferred. I therefore willingly adopted the suggestion of a friend, the benefit of whose advice I have on many occasions experienced, and made use of the following ordinary method of approximating to the root of the equation.
If the equation sin ξ - m sin (2 - θ) = 0 (see [page 274]) be regarded as an expression for the error, when the true value of ξ which would satisfy the equation has been introduced into its first member, we may consider any error in the value of ξ as expressed by the equation:
sin ξ - m . sin (2 ξ - θ) = ε
and differentiating this expression we have:
| d ε | = | cos ξ . d ξ - 2 m cos (2 ξ - θ) . d ξ |
| = | {cos ξ - 2 m cos (2 ξ - θ)} . d ξ |