\(\newcommand{L}[1]{\| #1 \|}\newcommand{VL}[1]{\L{ \vec{#1} }}\newcommand{R}[1]{\operatorname{Re}\,(#1)}\newcommand{I}[1]{\operatorname{Im}\, (#1)}\)

The angle sum rule

The angle sum rule is:

\[ \begin{align}\begin{aligned}\sin(\alpha \pm \beta) = \sin \alpha \cos \beta \pm \cos \alpha \sin \beta\\\cos(\alpha \pm \beta) = \cos \alpha \cos \beta \mp \sin \alpha \sin \beta\end{aligned}\end{align} \]

Proof

Let’s say we have a vector \((x_1, y_1)\) resulting from the anticlockwise rotation of a length 1 vector \((1, 0)\) by \(\alpha\) degrees around the origin.

We rotate this vector another \(\beta\) degrees anticlockwise around the origin to give length 1 vector \((x_2, y_2)\).

_images/angle_sum.png

We can see from the picture that:

\[ \begin{align}\begin{aligned}\cos(\alpha + \beta) = x_2 = r - u\\\sin(\alpha + \beta) = y_2 = t + s\end{aligned}\end{align} \]

We are going to use some basic trigonometry to get the lengths of \(r, u, t, s\).

Because the angles in a triangle sum to 180 degrees, \(\phi\) on the picture is \(90 - \alpha\) and therefore the angle between lines \(q, t\) is also \(\alpha\).

Remembering the definitions of \(\cos\) and \(\sin\):

\[ \begin{align}\begin{aligned}\cos\theta = \frac{A}{H} \implies A = (\cos \theta) H\\\sin\theta = \frac{O}{H} \implies O = (\sin \theta) H\end{aligned}\end{align} \]

Thus:

\[ \begin{align}\begin{aligned}p = \cos \beta\\q = \sin \beta\\r = (\cos \alpha) p = \cos \alpha \cos \beta\\s = (\sin \alpha) q = \sin \alpha \cos \beta\\t = (\cos \alpha) q = \cos \alpha \sin \beta\\u = (\sin \alpha) q = \sin \alpha \sin \beta\end{aligned}\end{align} \]

So:

\[ \begin{align}\begin{aligned}\cos(\alpha + \beta) = x_2 = r - u = \cos \alpha \cos \beta - \sin \alpha \sin \beta\\\sin(\alpha + \beta) = y_2 = t + s = \sin \alpha \cos \beta + \cos \alpha \sin \beta\end{aligned}\end{align} \]