# Helping Ada Lovelace with her Homework: Classroom Exercises from a Victorian Calculus Course – Mary Somerville’s Solution to the Trigonometry Problem

Author(s):

Somerville’s solution to the slightly harder Trigonometry Problem posed by Lovelace is contained in a letter, written on November 28, 1835 [LB 174, 28 Nov. 1835, ff. 31v-32r]. In it, Somerville writes:

The formulae proposed are
$\begin{array}{c}R \sin a=\sin⁡(a-b) \cos ⁡b + \cos⁡(a-b) \sin b\\ R \cos a=\cos⁡(a-b) \cos ⁡b - \sin⁡(a-b) \sin b\end{array}$

If the first be multiplied by $$\cos ⁡b$$, and the other by $$\sin ⁡b$$, their difference is

$R (\sin a \cos b - \cos a \sin b )= \sin⁡(a-b) ( \cos^2 ⁡b + \sin^2 b)$

but $$\cos^2 b+\sin^2 b=R^2$$, hence after dividing by $$R$$
$\sin ⁡a \cos ⁡ b - \cos ⁡a \sin ⁡b = R \sin⁡(a-b).$

Again, if the first be multiplied by $$\sin ⁡b$$ and the second by $$\cos ⁡b$$, their sum is

$R(\sin ⁡a \sin ⁡ b +\cos ⁡a \cos ⁡b ) = \cos⁡(a-b) ( \sin^2 ⁡b + \cos^2 b).$

Substituting $$R^2$$ for $$\sin^2 b+\cos^2 b$$, and then dividing by $$R$$ you will find
$\sin ⁡a \sin ⁡ b + \cos ⁡a \cos b = R \cos⁡(a-b).$