dave wrote:

You say QED - but is it? Maybe being pedantic here but is this a proof
as such? I mean what you have written is correct - but it's "just" a
particular case that happens to give the result expected. As I say,
not being pedantic but wonder if this really is a "proof"?

I'd say that if anything is missing it's that Roger didn't /prove/ that
the perpendicular from one corner of the equilateral triangle to the
opposite side actually bisects the angle and the opposite side. It's
obvious from the symmetry that it does, but can you prove it starting
only from Euclid's axioms?

I'm no mathematician but the definition of a cosine is, I believe, a
series.

Hmm... As usually taught you start with the simple right-angled
triangle definitions - cos = adjacent/hypotenuse, etc. - and then extend
that definition to allow angles outside the range 0 to 90 deg. Then the
power series expansions can be derived and the trig functions
generalised to allow complex number arguments. Hence if you can prove
the bisection of the angle and side above and accept Pythagoras (which
can be proved in several ways) you know that the 1 : root 3 : 2 triangle
has angles of 30, 60 & 90 deg. and that cos 30 deg. = sqrt(3)/2, and so on.

This was second form stuff when I was at school. It's a bit worrying to
read things like

My 1988 GCSE Maths didn't stretch to the relationships of Pythagoras
and trig. Why does sqrt(3)/2 = cos(30deg)?

(Discuss.)

--
Andy