I recently found out that $$\sin\frac\pi5=\frac12\sqrt{\frac{5-\sqrt5}2}$$ Which means that $$\cos\frac\pi5=\frac{1+\sqrt5}4$$ I also recently found that if $n\in\Bbb N$, $$\sin nx=\sin x\,U_{n-1}(\cos x)\\ \cos nx=T_n(\cos x)$$ Where $T_n$ and $U_n$ are the Chebyshev polynomials of the first and second kinds respectively. They are defined as $$T_n(x)=\frac{n}2\sum_{r=0}^{\lfloor n/2\rfloor}\frac{(-1)^n}{n-r}{n-r\choose r}(2x)^{n-2r}$$ $$U_n(x)=\sum_{r=0}^{\lfloor n/2\rfloor}(-1)^r{n-r\choose r}(2x)^{n-2r}$$ Using these definitions, I am attempting to find a general analytic expression for $\cos\frac\pi n$. To do so I started with $$\cos nx=T_n(\cos x)$$ $$\cos n\cdot\frac{\pi}{n}=T_n\bigg(\cos\frac{\pi}n\bigg)$$ then setting $w=\cos(\pi/n)$, $$T_n(w)+1=0$$ So our task is to solve for $w$. I'm fairly certain that $x=\cos(\pi/n)$ is always the largest real root of $T_n(x)+1=0$. So I guess that's my question:
What Is the largest real root of $T_n(x)+1=0$?
But as far as I know there are no really simple ways of going about this (but I really don't know much). Could I have a bit of help? Thanks.