find spectrum matrix A

Let $B_n = \frac{1}{\sqrt{bc}}(I_n - A_n)$ where $A_n$ is the matrix described above, and let $\psi_n(\lambda) = \det(B_n - \lambda I_n)$ be $B$'s characteristic polynomial. $B_n$ will be a tridiagonal matrix with the the values $-\frac{\sqrt{c}}{\sqrt{b}}, \frac{1}{\sqrt{bc}}, -\frac{\sqrt{b}}{\sqrt{c}}$ on each row.

If we expand $\det(B_n - \lambda I_n)$ by the last row we get $$\begin{align} \psi_n(\lambda) &= \left( \frac{1}{\sqrt{bc}} - \lambda \right) \psi_{n-1}(\lambda) - \psi_{n-2}(\lambda) \end{align}$$ and we have a recurrence starting condition that $\psi_1(\lambda) = \frac{1}{\sqrt{bc}} - \lambda$.

The above recurrence is strikingly similar to the recurrence for the Chebyshev polynomials of the second kind $U_n$, which are described by the recurrence: $$\begin{align} U_0(x) &= 1 \\ U_1(x) &= 2x \\ U_n(x) &= 2xU_{n-1}(x) - U_{n-2}(x) \end{align}$$ so if we change variables by setting $2x = \frac{1}{\sqrt{bc}} -\lambda$ and set $\psi_0(x) = 1$ we get the above recurrence: $$\begin{align} \psi_0(x) &= 1 \\ \psi_1(x) &= 2x \\ \psi_n(x) &= 2x\psi_{n-1}(x) - \psi_{n-2}(x) \end{align}$$

It is well-known that the roots of $U_n$ are: $$x_k = \cos \left( \frac{k}{n+1}\pi \right)$$ which gives us the eigenvalues for $B_n$: $$\lambda_k = \frac{1}{\sqrt{bc}} - 2x_k = \frac{1}{\sqrt{bc}} - 2\cos \left( \frac{k}{n+1}\pi \right)$$

Now, if $\lambda_k$ is an eigenvalue of $B_n$, what will the corresponding eigenvalue $\mu_k$ be for $A_n$? We have: $$\lambda_k v_k = B_n v_k = \frac{1}{\sqrt{bc}}\left(I_n - A_n \right)v_k = \frac{1}{\sqrt{bc}}(v_k - A_nv_k)$$ so $A_nv_k = v_k - \sqrt{bc} \lambda_k v_k = (1 - \sqrt{bc}\lambda_k)v_k$, i.e. $$\mu_k = 1 - \sqrt{bc}\lambda_k = 2 \sqrt{bc} \cos \left( \frac{k}{n+1}\pi \right) $$