Past exam of the mathematical methods I course of the University of Oxford / 2014 resit / 10 / a / i / Solution

For $n\ge 1$, use ${\sin }^2(nx)=[1-\cos (2nx)]/2$. The inner product isThe stated nonnegative index range also includes $n=0$. Since $f_0$ is the zero function, $⟨f_0,f_0⟩=0$, so it must be removed from any proposed basis.