I know that the coefficients that minimize the expression are the ones that make it's derivative 0. I have also expanded the whole expression and taken it's derivative, but still I can't figure out the coefficients. The solution must involve differentiating En. Any ideas?
For $1 \le k \le N$, $$ \begin{align} \frac{\partial E_{N}}{\partial b_{k}} & = -\int_{-\pi}^{\pi}2\left[f(x)-\sum_{n=1}^{N}b_{n}\sin nx\right]\sin kx\,dx \\ & = -2\int_{-\pi}^{\pi}f(x)\sin kxdx+2b_{k}\int_{-\pi}^{\pi}\sin^{2}kx\,dx \\ & = -2\int_{-\pi}^{\pi}f(x)\sin kx dx +2b_{k}\pi. \end{align} $$ Therefore, $$ \frac{\partial E_{N}}{\partial b_{k}} = 0 \iff b_{k}=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin kx dx. $$ The second derivative matrix is diagonal and equal to $2\pi I$.