Eigenvalues of a matrix close to a tridiagonal Toeplitz matrix 
I am trying to find all the eigenvalues of $P$ defined below:

$$P=\begin{bmatrix}
0.5&0.5&0&0&\cdots 0&0 \\
0.25&0.5&0.25&0&\cdots 0&0\\
0&0.25&0.5&0.25&\cdots 0&0\\
\vdots \\
0&0&0&0&\cdots 0.5&0.5
\end{bmatrix}_{n\times n}$$
So $P$ has $0.5$ along the main diagonal. It has $0.25$ on diagonals above and below the main diagonal except for the first and last row. Hence $P$ is not exactly a Toeplitz matrix.
My attempt: A paper I'm looking at, gives eigenvalues of the following Toeplitz matrix
$$
Q=\begin{bmatrix}
b&a \\
c&b&a \\
&\ddots&\ddots&\ddots \\
&&c&b&a \\
&&&c&b
\end{bmatrix}
$$
as $\lambda_j = b+2\sqrt{ca}\cos\left(\frac{j\pi}{n+1} \right)$
So I'm wondering if I will be able to find eigenvalues of $P$ even though its not a Toeplitz matrix precisely?
 A: $\def\a{\alpha}$
Let $u$ be an eigenvector of the matrix $P$ and assume the elements of the eigenvector have the form:
$$
u_k=e^{\alpha k}+ae^{-\alpha k},\tag1
$$
with some parameters $a$ and $\alpha$, which are to be found.
Obviously for all $k=2\dots(n-1)$
$$
(Pu)_k=\frac14\left(e^{\a k}+ae^{-\alpha k}\right)\left(e^\alpha+2+e^{-\alpha}\right)
=\frac14\left(e^\alpha+2+e^{-\alpha}\right)u_k.\tag2$$
Thus it remains only to find such $a$ and $\alpha$ that the equation $(2)$ is satisfied for $k=1$ and $k=n$ as well.
For $k=1$:
$$\begin{align}
&\frac12(e^{\a}+ae^{-\a}+e^{2\a}+ae^{-2\a})=\frac14\left(e^\a+2+e^{-\a}\right)\left(e^\alpha+a e^{-\alpha}\right)\\
&\iff
e^{2\a}+ae^{-2\a}=1+a \iff a=e^{2\a}.\tag{3}
\end{align}$$
For $k=n$:
$$\begin{align}
&\frac12(e^{\a(n-1)}+ae^{-\a(n-1)}+e^{\a n}+ae^{-\a n})
=\frac14\left(e^\a+2+e^{-\a}\right)\left(e^{\a n}+a e^{-\a n}\right)\\
&\iff e^{\a (n-1)}+ae^{-\a(n-1)}=e^{\a (n+1)}+ae^{-\a(n+1)}\\
&\iff e^{\a (n-1)}+ae^{-\a (n-1)}=e^{2\a}e^{\a(n-1)}+ae^{-2\a}e^{-\a(n-1)}\\
&\stackrel{(3)}\iff \left(e^{\a (n-1)}-e^{-\a (n-1)}\right)\left(1-e^{2\a}\right)=0\\
&\iff \a_m=\frac{\pi\,i}{n-1}m,\quad m=0\dots n-1
.\tag{4}
\end{align}$$
The corresponding eigenvalues are:
$$
\lambda_m=\frac14\left(e^{\a_m}+2+e^{-\a_m}\right)=\left(\frac{e^{\frac{\a_m}2}+e^{-\frac{\a_m}2}}2\right)^2=\cos^2\frac{\pi m}{2(n-1)},\quad m=0\dots n-1.\tag5
$$
Since all $n$ eigenvalues are distinct we are done.
A: (Not a solution)
I numerically found that the answer should be
$$
\lambda_j = \sin^{2}\left(\frac{j\pi}{2(n-1)}\right), \quad 0\leq j \leq n-1
$$
but I only have a vague idea to prove this. If we define $B_n = 2P_n - I_n$, then it is enough to show that the eigenvalues of $B_n$ are
$$
\cos\left(\frac{j\pi}{n-1}\right),\quad 0\leq j\leq n-1.
$$
It seems that the characteristic polynomial of $B_n$ is
$$
\phi_n(x) = \frac{1}{2^{n}}(x^2-1)U_{n-2}(x)
$$
where $U_n(x)$ is $n$-th Chebyshev polynomial of second kind. However, I'm not sure how to prove this, although it seems that using induction might give recurrence formula for $\phi_n$ which resembles that of $U_n$ a lot.
