matrices - eigenvalue problem of a simple circulant matrix

I've learn that the eigenvalues of this matrix
$$A=\begin{bmatrix}
2&-1&0&0&0\\
-1&2&-1&0&0\\
\vdots&\ddots&\ddots&\ddots&\vdots\\
0&0&-1&2&-1\\

0&0&0&-1&2
\end{bmatrix}$$
is $$\lambda_{j}=4\sin^2{\dfrac{j\pi}{2(n+1)}},j=1,2,\cdots,n$$
see How find this matrix has eigenvalues $\lambda_{j}=4\sin^2{\dfrac{j\pi}{2(n+1)}}$

And I think it is also possible to get similar eigenvalue result for a similar but circulant matrix, but I don't know how? any help? thanks in advance.
$$A=\begin{bmatrix}
2&-1&0&0&\color{red}{-1}\\
-1&2&-1&0&0\\
\vdots&\ddots&\ddots&\ddots&\vdots\\

0&0&-1&2&-1\\
\color{red}{-1}&0&0&-1&2
\end{bmatrix}$$