How prove this $\cot(\pi/15)-4\sin(\pi/15)=\sqrt{15}$ I need some help with this demonstration, please
I have tried with some identities but nothing.
I wanted to use this $$\sin(\pi/15)\cdot \sin(2\pi/15)\cdots\sin(7\pi/15)=\sqrt{15}$$

We may prove:
$$ \cos\frac{\pi}{15}-4\sin^2\frac{\pi}{15}=\sqrt{15}\sin\frac{\pi}{15} $$
by squaring both sides. By setting $\theta=\frac{\pi}{15}$, that leads to:
$$ \frac{13}{2}-2\cos(\theta)-\frac{15}{2}\cos(2\theta)+2\cos(3\theta)+2\cos(4\theta) = \frac{15}{2}-\frac{15}{2}\cos(2\theta)$$
or to:
$$ -\cos(\theta)+\cos(3\theta)+\cos(4\theta) = \frac{1}{2} $$
so we just have to prove that $\cos(\theta)$ is a root of:
$$ p(x) = 16x^4+8x^3-16x^2-8x+1.$$
That easily follows from:
$$ \Phi_{30}(x) = x^8+x^7-x^5-x^4-x^3+x+1.$$
