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.
$$A=\begin{bmatrix} 2&-1&0&0&\color{red}{-1}\\ -1&2&-1&0&0\\ \vdots&\ddots&\ddots&\ddots&\vdots\\ 0&0&-1&2&-1\\ \color{red}{-1}&0&0&-1&2 \end{bmatrix}$$