Andrei from Romaniaproved the double angle formulae
illustrated in the diagram:
The diagram starts from a right angled triangle,
of sides 2t and 2 and where consequently
tanq = t. In this triangle, a line making
an angle q with the hypotenuse is drawn.
This way, an isosceles triangle is formed, and
2q is the angle exterior to this isosceles
triangle. Let the sides DA and DB of this
isosceles triangle be x units. Then the length
of DC must be 2-x units. Using Pythagoras
Theorem for triangle ADC we find x.
x2=(2t)2+(2-x)2.
Hence x=1+t2 and so the length of side DC is
2-(1+t2)=1-t2.
The formulae for the sine, cosine and tangent of
2q in terms of t, where t=tanq,
follow directly from the ratios of the sides of
the right angled triangle ADC and we get