Q:

How can I prove $\tan15^\circ=2-\sqrt3$ using the triple-angle formulas?

Prove $\tan15^\circ=2-\sqrt3\,\,$ using following results $$\sin3x=3\sin{x}-4\sin^3x$$
  $$\cos3x=4\cos^3x-3\cos{x}$$

I know that if I substitute $x=15^\circ\,$I can write$$\sin45=3\sin15-4\sin^315$$
$$\cos45=4\cos^315-3\cos15$$ What can I do next? any hints?

A:

Let $s=\sin 15^\circ$, $c=\cos 15^\circ$ and  $t=s/c=\tan 15^\circ\in (\tan 0^\circ,\tan 45^\circ)=(0,1)$ then, from your equations, since $\sin 45^\circ=\cos 45^\circ$, it follows that
$$s(3-4s^2)=c(4c^2-3)\implies s(3(s^2+c^2)-4s^2)=c(4c^2-3(s^2+c^2))$$
and, after dividing both sides by $c^3$, we obtain
$$t(3-t^2)=(1-3t^2)\implies (t+1)(t^2-4t+1)=0.$$
(the factorization can be obtained by noticing that the equation on the left is satisfied by $t=-1$).
Can you take it from here and find the root $t\in (0,1)$?

