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

educationtrigonometry

In High School, the above mentioned formula is often proved assuming that $x,y$ are acute angles and using some more or less 'brilliant' geometrical construction.

And after that, books 'forget' the assumption about the angles and apply the identity to whatever angles.

There is a trivial, long and tedious proof by cases, with all combinations of the possible quadrants where $x$ and $y$ can lie.

Is there a shorter way to extend the formula for whatever angles? The proof must be suitable for High School, that is, no Taylor series or Euler's formula.

I don't think that this is a duplicate of the proposed question, because the first answer considers only acute angles and the second uses Euler's formula.

Best Answer

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.