Is there a short, elementary way to show $\cos(x+y)=\cos x\cos y-\sin x\sin y$ for any angles $x$ and $y$?

I learned this proof from this wonderful 1-page paper Proof of Sum and Difference Identities by Gilles Cazelais

Two black outlined circles with radius 1 is laid out side by side, each centered on x and y axes against a plain white background. Each circle has an outlined triangle inside with one corner in the center, and the other two on the outline of the circle of which the distance between is labelled d. On the left, both edges of the triangle from the center do not coincide with any of the axes; the counterclockwise angle from the x-axis of one of those edges is labelled alpha and the other beta. On the right, one of the edges coincide with the x-axis, the other making a counterclockwise angle from the x-axis of alpha minus beta. The corners of the triangles meeting the outline of the circle have their coordinates labelled as well, which are written in terms of cos, sin and their respective angles.

$$ d = \sqrt{(\cos\alpha - \cos\beta)^2 + (\sin\alpha - \sin\beta)^2} \\ d= \sqrt{(\cos(\alpha-\beta)-1)^2+(\sin(\alpha-\beta)-0)^2} $$

from which you could even leave as an exercise for high school students to prove.


Durell & Robson's proof seems satisfactory to me. It refers back to their Elementary Trigonometry, Part III, "The General Angle and Compound Angles", for proofs that:

\begin{align*} \cos\left(A + \frac\pi2\right) & = -\sin A, \\ \sin\left(A + \frac\pi2\right) & = \cos A, \end{align*} for values of $A$ of any magnitude, positive or negative.

Unfortunately, my copy of E.T. doesn't contain Part III.

[Added later: I found a PDF copy. It gives Euclid-style proofs, by congruent triangles, with figures for the cases where $A$ is in the second or third quadrant. The reader is asked to supply figures for the cases of the first and fourth quadrants.]

It still seems worth typing out their short proof of the addition formulae from p.123f. of Advanced Trigonometry (1930, Dover reprint 2003):

Let the directed lines $O\xi, OP, O\eta$ make angles $A, A+B, A + \frac\pi2$ with $Ox$; and let the projections of $P$ on $O\xi, O\eta$ be $N, M$. Suppose that $OP$ contains $l$ units of length.

The positions of $N, M$ on the directed lines $O\xi, O\eta$ are given by the directed numbers which measure $ON, OM$, and these are, by the definitions of the cosine and sine of the general angle, $l\cos B, l\sin B$.

$\therefore$ by [an earlier equation], \begin{align*} \text{Projection of } ON \text{ on } Ox & = l\cos B \cdot\cos A, \\ \text{Projection of } OM \text{ on } Ox & = l\sin B \cdot\cos\left(A + \frac\pi2\right). \\ \text{Also the projection of } OP \text{ on } Ox & = l\cos(A+B). \end{align*}

But the projection of $OP$ on $Ox$ is equal to the sum of the projections on $ON, NP$, i.e. to the sum of the projections of $ON, OM$ on $Ox$. $$ \therefore\ l\cos(A+B) = l\cos B\cos A + l\sin B \cdot\cos\left(A + \frac\pi2\right). $$ But $\cos\left(A + \frac\pi2\right) = -\sin A$, see E.T., pp. 199, 200; $$ \therefore\ \cos(A+B) = \cos A\cos B - \sin A\sin B. $$ Further, if the directed line $Oy$ makes $+\frac\pi2$ with $Ox$, the projections of $ON, OM, OP$ on $Oy$ are $$ l\cos B\sin A, \ l\sin B\sin\left(A + \frac\pi2\right), \ l\sin(A+B). $$ $\therefore$ as before, $$ l\sin(A+B) = l\cos B\sin A + l\sin B \cdot\sin\left(A + \frac\pi2\right). $$ But $\sin\left(A + \frac\pi2\right) = \cos A$, see E.T., pp. 199, 200; $$ \therefore\ \sin(A+B) = \sin A\cos B + \cos A\sin B. $$ This proof holds good for values of $A$ and $B$ of any magnitude, positive or negative. Figs 59, 60 [omitted] show two possible cases; the reader should draw other figures (e.g. $A = 100^\circ$, $B = 50^\circ$ or $A = 220^\circ$, $B = 160^\circ$) and satisfy himself that the proof applies to them, without any modification.

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