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 $\tan \theta =t$. In this triangle, a line making an angle $\theta$ with the hypotenuse is drawn. This way, an isosceles triangle is formed, and $2\theta$ 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 $x$.

${\displaystyle x^2=(2t{)}^2+(2-x{)}^2.}$

Hence $x=1+t^2$ and so the length of side DC is $2-(1+t^2)=1-t^2$.

The formulae for the sine, cosine and tangent of $2\theta$ in terms of $t$, where $t=\tan \theta$, follow directly from the ratios of the sides of the right angled triangle ADC and we get

${\displaystyle \tan 2\theta =\frac{2t}{1-t^2},\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}\sin 2\theta =\frac{2t}{1+t^2},\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}\cos 2\theta =\frac{1-t^2}{1+t^2}}$