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 $$\sin(\pi/15)\cdot \sin(2\pi/15)\cdots\sin(7\pi/15)=\sqrt{15}$$
 A: 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:
$$ \Phi_{30}(x) = x^8+x^7-x^5-x^4-x^3+x+1.$$
A: Consider the equation $$\tan5\theta=\tan\dfrac{\pi}{3}$$ which has $5$ principle solutions $$\theta=\dfrac{n\pi}{5}+\dfrac{\pi}{15}\,\,\,\,\,\,\,\,n=0, 1, 2, 3, 4.$$ Therefore $\tan\dfrac{\pi}{15}$ is a root of the equation $$t^5-5\sqrt3t^4-10t^3+10\sqrt3t^2+5t-\sqrt3=0.$$ If we can find $\tan\dfrac{\pi}{15}$ using above equation we just have to show that $$\dfrac{1}{t}-\dfrac{4t}{\sqrt{t^2+1}}=\sqrt{15}.$$ After solve the above equation, I got  that $$\color{Green}{\tan\dfrac{\pi}{15}=\sqrt{23-10\sqrt5-2\sqrt{255-114\sqrt5}}=\dfrac{3\sqrt3}{2}-\dfrac{\sqrt{15}}{2}-\dfrac{\sqrt{50-22\sqrt5}}{2}}.$$
