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Andrei from Romania proved 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 $x$. $$x^2=(2t)^2+(2-x)^2.$$ 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 $$\tan2\theta = {2t\over {1-t^2}},\quad \sin2\theta = {2t\over {1+t^2}},\quad \cos2\theta = {{1-t^2}\over {1+t^2}}$$