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

Adrian Rice (Randolph-Macon College)

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

Return to Main Solutions Page.
Continue to Solution to Problem 1.