How to find the eigenvalues of tridiagonal Toeplitz matrix? Assume the tridiagonal matrix $T$ is in this form:
$$
T = \begin{bmatrix}
a       & c        &              &              &   &\\
b       & a        & c            &              &   &\\
        & b        & a            & c            &   &\\
        &          &              &\ddots        &   &\\
        &          &              & b            & a & c\\
        &          &              &              & b & a\\
\end{bmatrix}
$$
we must show that its eigenvalues are of the form 
$$a + 2 \sqrt{bc} \, \cos \left( \frac{k \pi}{n+1} \right)$$
where $$a=qh^2−1, ~~ b=1- \frac{ph}{2}, ~~ c =1+\frac{ph}{2} , ~~q \leq 0.$$

Developing the determinant on the first column, then the second minor on its first column, you obtain the key recurrence
$$\Delta_n=a\Delta_{n-1}-bc\Delta_{n-2},$$
with $\Delta_0=1$ and $\Delta_1=a$.
The solution of the recurrence is
$$\Delta_n=\frac{\left(\dfrac{a+\sqrt{a^2-4bc}}2\right)^{n+1}-\left(\dfrac{a-\sqrt{a^2-4bc}}2\right)^{n+1}}{\sqrt{a^2-4bc}}.$$
This expression cancels when
$$\frac{a-\sqrt{a^2-4bc}}{a+\sqrt{a^2-4bc}}=\frac{1-\sqrt{1-\dfrac{4bc}{a^2}}}{1+\sqrt{1-\dfrac{4bc}{a^2}}}$$ is an $(n+1)^{th}$ root of one, let $\omega_k$.
Then
$$\frac{4bc}{a^2}=1-\left(\frac{\omega_k-1}{\omega_k+1}\right)^2=\frac{2\omega_k}{(\omega_k+1)^2}$$ and
$$a=\pm\frac{\omega_k+1}{2\sqrt{\omega_k}}2\sqrt{bc}.$$
The coefficient in $\omega_k$ is real and can be written
$$\cos\frac{k\pi}{n+1}.$$
Then with $a-\lambda$ instead of $a$, we have established
$$\lambda= a\pm\cos\frac{k\pi}{n+1}2\sqrt{bc}.$$
