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 $t\in (0,1)$?
A: Solution
For convenience, denote $\sin 15^o=x, ~~\cos 15^o=y.$ Obviously, $0<x<y$ and $x^2+y^2=1$.
Thus, we obtain $$3x-4x^3=\frac{\sqrt{2}}{2},\tag1$$ and $$4y^3-3y=\frac{\sqrt{2}}{2}. \tag2$$
By $(1)-(2)$, we have $$3x+3y-4x^3-4y^3=(x+y)[3-4(x^2-xy+y^2)]=0.\tag3$$
Thus, $$x^2-xy+y^2=\frac{3}{4}.\tag4$$
Hence $$xy=\frac{1}{4}.\tag5$$
Further,$$(x+y)^2=x^2+y^2+2xy=1+2\cdot \frac{1}{4}=\frac{3}{2}.\tag6$$
Hence,$$x+y=\frac{\sqrt{6}}{2}.\tag7$$
From $(5),(7)$, we may see that $x, y$ are the roots for the quadratic equation $$t^2-\frac{\sqrt{6}}{2}t+\frac{1}{4}=0.\tag8$$
By solving it, we have $$x=\frac{\sqrt{6}-\sqrt{2}}{4},~~y=\frac{\sqrt{6}+\sqrt{2}}{4}.$$
As a result, $$\tan 15^o=\frac{x}{y}=2-\sqrt{3}.$$
Note
Now,you must have seen that the algebraic method is too complicated. Indeed, there exists an elegant geometrical solution without much calculation and even without a word. Can you get the same result from here?

