Past exam of the mathematical methods I course of the University of Oxford / 2015 resit / 12 / a

For a positive integer $n$, the identity ${\sin }^2(nx)=[1-\cos (2nx)]/2$ givesThe stated nonnegative indexing includes $n=0$ as a special case: $f_0\equiv 0$, whose squared norm is zero. Thus $⟨f_0,f_0⟩=0$; it cannot belong to a basis.

For distinct positive integers $m,n$, the product-to-sum identity givesBoth cosine integrals vanish over $[-\pi ,\pi ]$, soIf one index is zero, the corresponding function is identically zero and the same answer follows immediately.