Orthonormality of $\{1,\cos nt,\sin (n-a)t\}_{n\in\mathbb N}$.

Question

We know that the system $$\{1,\cos nt,\sin nt\}_{n\in\mathbb N}$$ is orthonormal on $[0,2\pi]$ respect to $\frac{1}{\pi}\int_{0}^{2\pi} \cdot\ dt$. My question is: For what values of the parameter $a\in\mathbb R$ the system $$\{1,\cos nt,\sin (n-a)t\}_{n\in\mathbb N}$$ remains orthonormal?

Answer 1

Let $k\in \mathbb{Z}$. First we have \begin{align} \int_0^{2\pi}\sin^2[(n-a)t]dt=\pi-\frac{\sin[4(a-n)\pi]}{4(a-n)}, \end{align} hence we need $\sin[4(a-n)\pi]=0$ or that $4(a-n)\pi=k\pi$, thus $a=\frac k4+n$, hence a need not necessarily be an integer. \begin{align} \int_0^{2\pi}\sin[(n-a)t]dt=-2\frac{\sin^2[(a-n)\pi]}{a-n}, \end{align} thus we need $(a-n)\pi=k\pi$ or that $a=k+n$, i.e. $a\in \mathbb{Z}$, hence $a$ must be an integer.

Last \begin{align} \int_0^{2\pi}\cos[nt]\sin[(n-a)t]dt=\frac{-2a+2n+(a-2n)\cos[2a\pi]+a\cos[2(a-2n)\pi]}{2a(a-n)}, \end{align} thus we need $\cos[2a\pi]=1$ and $\cos[2(a-2n)\pi]=1$, implying that $a\in \mathbb{Z}$, hence $a$ must be an integer.

Putting all these together we see that $a$ can be any integer. Note that the cases where $a=0,2n,n$ are trivial.