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

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

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