Let $A\in\mathbf{R}^{NxN}$ be a matrix of the form
\begin{pmatrix} 2 & -1 & 0 & \cdots & \cdots & \cdots & \cdots & 0\\ -1 & 2 & -1 & 0 & & & & \vdots\\ 0 & -1 & 2 & -1 & \ddots & & & \vdots\\ \vdots & 0 & \ddots & \ddots & \ddots & \ddots & & \vdots\\ \vdots & & \ddots & \ddots & \ddots & \ddots & 0 & \vdots\\ \vdots & & & \ddots & -1 & 2 & -1 & 0\\ \vdots & & & & 0 & -1 & 2 & -1\\ 0 & \cdots & \cdots & \cdots & \cdots & 0 & -1 & 2\\ \end{pmatrix}
My course says that the eigenvalues and the eigenvectors of $A$ are given by: $$\lambda_{j}= -2+2cos(\frac{j\pi}{N+1}).$$ $$ v_j=(sin(\frac{j\pi}{N+1}),..., \frac{Nj\pi}{N+1}))^t.$$
I suspect that $v_j^t v_j=constant*\delta_{i,j}$. However, I have been unable to prove this and was wondering if anybody might be able to help me.
For $2\le k\le N-1$ we have the linear difference equation $-v_{k-1}^{(j)}+(2-\lambda)v_k^{(j)}-v_{k+1}^{(j)}=0$. Substitute $v_k^{(j)}=A^k$ and we get the characteristic equation $A^2+(\lambda-2)A+1=0$ with solutions $$A=\frac{2-\lambda+\sqrt{(2-\lambda)^2-4}}2$$ $$A^{-1}=\frac{2-\lambda-\sqrt{(2-\lambda)^2-4}}2$$ So the general solution is $v_k^{(j)}=c_1A^k+c_2A^{-k}$.
Apply the boundary conditions $$(2-\lambda)\left(c_1A+c_2A^{-1}\right)-\left(c_1A^2+c_2A^{-2}\right)=0$$ and $$-\left(c_1A^{N-1}+c_2A^{-(N-1)}\right)+(2-\lambda)\left(c_1A^n+c_2A^{-N}\right)=0$$ And we get $$\begin{align}\frac{c_1}{c_2}&=-\frac{A^{-2}+(\lambda-2)A^{-1}}{A^2+(\lambda-2)A}=-A^{-2}\frac{A^{-1}+\lambda-2}{A+\lambda-2}\\ &=-\frac{A^{-(N-1)}+(\lambda-2)A^{-N}}{A^{N-1}+(\lambda-2)A^N}=-A^{-2N}\frac{A+\lambda-2}{A^{-1}+\lambda-2}\end{align}$$ Since $A+A^{-1}=2-\lambda$ we have $$-A^{-2}\frac{(-A)}{(-A^{-1})}=-A^{-2N}\frac{(-A^{-1})}{(-A)}$$ So $$A^{2(N+1)}=1=e^{2\pi ij}$$ And $$A=e^{\frac{\pi ij}{N+1}}=\cos\frac{\pi j}{N+1}+i\sin\frac{\pi j}{N+1}=\frac{2-\lambda+i\sqrt{4-(2-\lambda)^2}}2$$ From the real pat we see that $\lambda=2-2\cos\frac{\pi j}{N+1}$ and then the imaginary part checks. And since we have already shown that $c_1/c_2=-1$, we have $$v_k^{(j)}=c_2\left(-e^{\frac{\pi ijk}{N+1}}+e^{-\frac{\pi ijk}{N+1}}\right)=-2ic_2\sin\frac{\pi jk}{N+1}$$ This should be somewhere in the literature about the DST-1 but I'm not very good at searching the literature.
EDIT: I didn't see the part about orthogonality and normalization. First off, $$\sin\alpha\left(k+\frac12\right)-\sin\alpha\left(k-\frac12\right)=2\cos\alpha k\sin\frac{\alpha}2$$ So if $\sin\frac{\alpha}2\ne0$, then $$\begin{align}\sum_{k=1}^N\cos\alpha k&=\frac1{2\sin\frac{\alpha}2}\sum_{k=1}^N\left(\sin\alpha\left(k+\frac12\right)-\sin\alpha\left(k-\frac12\right)\right)\\ &=\frac{\sin\alpha\left(N+\frac12\right)-\sin\left(\frac{\alpha}2\right)}{2\sin\frac{\alpha}2}\end{align}$$ So if $j_1\ne j_2$ then $$\begin{align}\sum_{k=1}^Nv_k^{(j_1)}v_k^{(j_2)}&=\sum_{k=1}^N\sin\frac{\pi j_1k}{N+1}\sin\frac{\pi j_2k}{N+1}\\ &=\frac12\sum_{k=1}^N\left(\cos\frac{\pi(j_1-j_2)k}{N+1}-\cos\frac{\pi(j_1+j_2)k}{N+1}\right)\\ &=\frac{\sin\frac{\pi(j_1-j_2)(2N+1)}{2(N+1)}-\sin\frac{\pi(j_1-j_2)}{2(N+1)}}{4\sin\frac{\pi(j_1-j_2)}{2(N+1)}}-\frac{\sin\frac{\pi(j_1+j_2)(2N+1)}{2(N+1)}-\sin\frac{\pi(j_1+j_2)}{2(N+1)}}{4\sin\frac{\pi(j_1+j_2)}{2(N+1)}}\\ &=\frac{-(-1)^{j_1-j_2}\sin\frac{\pi(j_1-j_2)}{2(N+1)}-\sin\frac{\pi(j_1-j_2)}{2(N+1)}}{4\sin\frac{\pi(j_1-j_2)}{2(N+1)}}-\frac{-(-1)^{j_1+j_2}\sin\frac{\pi(j_1+j_2)}{2(N+1)}-\sin\frac{\pi(j_1+j_2)}{2(N+1)}}{4\sin\frac{\pi(j_1+j_2)}{2(N+1)}}\\ &=-\frac{(-1)^{j_1+j_2}+1}4+\frac{(-1)^{j_1+j_2}+1}4=0\end{align}$$ On the other hand, $$\begin{align}\sum_{k=1}^N\left(v_k^{(j)}\right)^2&=\sum_{k=1}^N\sin^2\frac{\pi jk}{N+1}=\frac12\sum_{k=1}^N\left(1-\cos\frac{2\pi jk}{N+1}\right)\\ &=\frac N2-\frac1{4\sin\frac{\pi j}{N+1}}\left(\sin\frac{\pi j(2N+1)}{N+1}-\sin\frac{\pi j}{N+1}\right)\\ &=\frac N2-\frac{-\sin\frac{\pi j}{N+1}-\sin\frac{\pi j}{N+1}}{4\sin\frac{\pi j}{N+1}}=\frac{N+1}2\end{align}$$ Now, you really knew all about this edit in the first place because eigenvectors of a Hermitian operator belonging to different eigenvalues are always orthogonal and the average value of $\sin^2\theta$ taken all the way around the circle is $\frac12$ so $$\frac12(2N+2)=\sum_{k=0}^{2N+1}\sin^2\frac{\pi jk}{N+1}=2\sum_{k=1}^N\sin^2\frac{\pi jk}{N+1}$$