How to show this matrix has eigenvalues $\lambda_{j}=4\sin^2{\frac{j\pi}{2(n+1)}}$? Show that the $n\times n$ tridiagonal matrix 
$$A=\begin{bmatrix}
2&-1&0&0&0\\
-1&2&-1&0&0\\
\vdots&\ddots&\ddots&\ddots&\vdots\\
0&0&-1&2&-1\\
0&0&0&-1&2
\end{bmatrix}
$$
has the eigenvalues
$$\lambda_{j}=4\sin^2{\dfrac{j\pi}{2(n+1)}},j=1,2,\cdots,n$$
 A: So we need to solve the following problem: $Ax =\lambda x$.
 Or $$cx_{j−1} + ax_j + bx_{j+1} = \lambda x_j , \ j = 1,\ldots ,m$$
$$x_0 = x_{m+1} = 0$$
where $a=2, \ b=c=-1$
which is equivalent to
$$cx_{j−1} +(a-\lambda)x_j + bx_{j+1} = 0 , j = 1,\ldots ,m$$
$$x_0 = x_{m+1} = 0$$
Its solution is represented in the form:
$$x_j = \alpha r_1^j+\beta r_2^j$$
where $r_1^j, \ r_2^j$ are solutions of the characteristic polymonial
$f(r)=b r^2+(a-\lambda)r+c$:
$$r_1=\frac{-(a-\lambda)+\sqrt{(a-\lambda)^2-4b c}}{2b}$$
$$r_2=\frac{-(a-\lambda)-\sqrt{(a-\lambda)^2-4b c}}{2b}$$
We can find the unknown coefficients by using the initial condition:
$$x_0 = \alpha+\beta = 0 \Rightarrow \alpha=-\beta$$
$$x_j = \alpha (r_1^j-r_2^j)$$
$$x_{m+1} = \alpha(r_1^{m+1}-r_2^{m+1})=0 \Rightarrow \left(\frac{r_1}{r_2}\right)^{m+1}=1$$
$$r_1 r_2=\frac{c}{b}$$
$$\left(\frac{r_1}{r_2}\right)^{m+1}=\left(\frac{b r_1^2}{c}\right)^{m+1}=1$$
Plugging in $b=c=-1$ leads us to $r^{m+1}=1 $, so 
$$r_{1,j}=e^{\pi i \left(\frac{j}{n+1}\right)}$$
$$r_{2,j}=e^{-\pi i \left(\frac{j}{n+1}\right)}$$
But one can see that $r_{1,j}+r_{2,j}=\frac{\lambda_{j}-a}{b}=2-\lambda_{j}$. Using the Euler's formula one can obtain:
$$\lambda_{j}=2\left(1-\cos\left(\frac{j\pi}{n+1} \right)\right)=4\sin^2\left(\frac{j\pi}{2(n+1)} \right)$$
A: Hint: Find $\det (A-\lambda I)$ by induction on size of $A$ (you will get a recurrent equation of order 2).
