I would like to find the roots of the polynomial sequence given by a recurrence relation as follows:

$V_0(x) = 1-a^2$

$V_1(x) = 1-a^2 - x$

$V_{k \geq 2}(x) = (1+a^2 - x)V_{k-1}(x) - a^2V_{k-2}(x)$

I know that $a \in (0,1)$.

Another formulation of the problem could be to find the eigenvalues of the tridiagonal symmetric matrix $C_n$

$C_n = \left[ \begin{matrix} 1-a^2 & -a\sqrt{1-a^2} & & \\ -a\sqrt{1-a^2} & 1+a^2 & -a & \\ & -a & 1+a^2 & \ddots \\ & & \ddots & \ddots \end{matrix} \right]$

Since the matrix is symmetric, the roots are going to be real. I also know that the matrices $C_n$ are positive definite.

This is almost like a Chebyshev recursion, but a little bit perturbed.

**A bonus question**: I have some vague memories about the proof of the roots of Chebyshev polynomials that involved an argument that the roots of the consecutive polynomials in the sequence separate each other, and from this, with some additional tools the roots were derived. If someone can point to a location where I can find that proof, it would be really helpful.

**Edit after the solutions: Based on the trigonometric equations derived by Pietro Mejer, I suspect that there is no closed formula that describes the roots.**