<s>[INST] 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 $$a=qh^2−1, ~~ b=1- \frac{ph}{2}, ~~ c =1+\frac{ph}{2} , ~~q \leq 0.$$
 [/INST] Consider the matrix $S$ which is equal to $T$ except that the diagonal is zero. That is, $T=a I + S$. If we knew the eigenvalues $\lambda_1,\ldots,\lambda_n$ of $S$, the eigenvalues of $T$ are simply $a+\lambda_1,\ldots,a+\lambda_n$.
Now let 
$$
D=\mathrm{diag}(\delta_1,\ldots,\delta_n), \quad \delta_i=\left(\frac{b}{c}\right)^{\frac{i-1}{2}},
$$
and notice that the matrix $P=D^{-1}SD$ is symmetric.
Indeed,
$$
p_{i+1,i}=\delta_{i+1}^{-1}b\delta_i=b\left(\frac{b}{c}\right)^{\frac{i}{2}+\frac{i-1}{2}}=\sqrt{bc},
\quad
p_{i,i+1}=\delta_i^{-1}c\delta_{i+1}=\cdots=\sqrt{bc}.
$$
Hence
$$
T=\alpha I + S \sim \alpha I + P = \alpha I + \sqrt{bc} Q,
$$
where $Q$ is zero everywhere except the ones in above and below the main diagonal and hence the eigenvalues of $T$ are $\alpha+\sqrt{bc}\mu$, where $\mu$ is an eigenvalue of $Q$.
Now assume that a vector $v=(v_i)$ is an eigenvector of $Q$ associated with the eigenvalue $\mu$. We have
$$\tag{1}
v_{i-1}+v_{i+1}=\mu v_i, \quad i=2,\ldots,n-1, \quad v_2 = \mu v_1, \quad v_{n-1} = \mu v_n.
$$
Assume $v_i$ in the form $v_i=\alpha\sin i\theta+\beta\cos i\theta$.
We have
$$
\begin{split}
v_{i-1}+v_{i+1}
&=
\alpha[\sin(i-1)\theta+\sin(i+1)\theta]+\beta[\cos(i-1)\theta+\cos(i+1)\theta]
\\&=
\alpha[\sin i\theta\cos\theta-\cos i\theta\sin\theta+\sin i\theta\cos\theta+\cos i\theta\sin\theta]
\\&+
\beta[\cos i\theta\cos\theta+\sin i\theta\sin\theta+\cos i\theta\cos\theta-\sin i\theta\sin\theta]
\\&=
2\alpha\sin i\theta\cos\theta
+
2\beta\cos i\theta\cos\theta
=2\cos\theta v_i.
\end{split}
$$
This yields $\mu=2\cos\theta$. It remains to find $\theta$; for this we use the boundary conditions in (1). So with $v_i=\alpha\sin\theta+\beta\cos\theta$,
$$
v_2=\mu v_1\quad\Leftrightarrow\quad \alpha\sin 2\theta+\beta\cos 2\theta=2\cos\theta(\alpha\sin\theta+\beta\cos\theta),
$$
$$
v_{n-1}=\mu v_n\quad\Leftrightarrow\alpha\sin(n-1)\theta+\beta\cos(n-1)\theta=2\cos\theta(\alpha\sin n\theta+\beta\cos n\theta).
$$
This gives a system for $\alpha$ and $\beta$:
$$
A\begin{bmatrix}\alpha\\\beta\end{bmatrix}=0,
$$
where
$$
\begin{split}
A&=\begin{bmatrix}
\sin 2\theta-2\sin\theta\cos\theta&\cos 2\theta-2\cos^2\theta\\
\sin(n-1)\theta-2\sin n\theta\cos\theta&\cos(n-1)\theta-2\cos n\theta\cos\theta
\end{bmatrix}
\\&=
\begin{bmatrix}
0 & -1 \\
-\sin(n+1)\theta & *
\end{bmatrix}.
\end{split}
$$
The first equation gives $\beta=0$. We look for a nonzero $\alpha$; for this we must have $\sin(n+1)\theta=0$.
This gives
$$
\theta:=\theta_k=\frac{k\pi}{n+1}, \quad \mu_k=2\cos\frac{k\pi}{n+1}.
$$
Hence the eigenvalues of $T$ are
$$
a+\sqrt{bc}\mu_k=a+2\sqrt{bc}\cos\frac{k\pi}{n+1}, \quad k = 1,\ldots,n.
$$
</s>