Once upon a time, when I was deep into trigonometry, I followed the trail of sine and cosine sums, with sin(A + B) = sin(A)cos(B) + cos(A)sin(B).
Then two more:
sin(A) + sin(B) = 2sin((A + B)/2)cos((A – B)/2)
and 2sin(A)cos(B) = sin(A + B) + sin(A – B)
Being unable to remember these last formulae (there are 8 altogether) I learned the basic one and derived, over and over again, the others.
In all of this I was never able to derive the basic formula until de Moivre’s theorem appeared.
So I had another go with trig and the simple version, and this is the result:
Aim of the game : sin(2x) = 2sin(x)cos(x)
First diagram:This does not look promising !
But what about sensible labelling –
The vertical from F meets AD in J
and the line from F at right angles to AB meets AB in K.
So we have two pairs of congruent triangles, BH and HD are both equal to sin(x),
and BHD is A STRAIGHT LINE
Now the vertical from B is sin(2x)
so sin(2x)/(2sin(x)) is the ratio which should be cos(x)