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.