Prove that the Chebyshev polynomials have interlacing zeros. Between any two neighboring zeros of $T_n$ there is precisely one zero of $T_{n-1}$. Prove this property. Where $$T_n(t) = \cos(n\theta), \,\,\,\theta = \arccos(t),\,\,\,\,n=0,1,2,\dots$$ I have the recursion formula $$T_{n+1}(t) = 2tT_n(t)-T_{n-1}(t)$$ and I have been to show this by contradiction by using the following set up. Let $a,b,c\in[-1,1]$ where $T_n(a)=0=T_n(b)$ and $a<c<b$. Assume that $T_{n-1}(c)\ne 0$. I have tried messing around with the recursion formula and getting nowhere. Any suggestions would be greatly appreciated!

Why don't you find the zeros of $T_n$ explicitly in terms of $\theta$? This is very simple to do.

Or if you do not want to do that, then you can just pick any two successive zeros of $T_n$, say $a$ and $b$, $a