I'm trying to prove that $ \big\{\sqrt{2}\cos\big((\pi/2+\pi n)x\big)\big\}_{n\in \mathbb{N}\cup\{0\}}$ is an orthonormal basis for $L^2[0,1]$.
I thought of using the fact that $ \big\{\sqrt{2}\cos(\pi nx)\big\}_{n\in \mathbb{N}\cup\{0\}}$ is an orthonormal basis for $L^2[0,1]$ by using the representation $f(x)=\sum_{n=1}^\infty\alpha_n \cos(\pi nx)$ to transition to the new basis and show that, it too, can yield $f(x)$, but there doesn't seem to be a trigonometric identity that helps accomplish this.
It seems redundant to prove from scratch that $ \big\{\sqrt{2}\cos\big((\pi/2+\pi n)x\big)\big\}_{n\in \mathbb{N}\cup\{0\}}$ is a basis for $L^2[0,1]$ given that there's one very similar. In any case, doing this from scratch, I have been able to show that this is an orthonormal system, but couldn't complete the proof to show it is a basis.
I've also been unsuccessful in trying to refute that it is a basis.
Any ideas either way?
A related post that may be helpful (but didn't help me) is this.
Let $f:(0,1) \rightarrow \mathbb{R}$ and $\frac{e^{i\pi x/2}}{2} g(x)$ its even extension on $(-1,1)$. Then $g(x)$ is some complex function on $(-1,1)$ which can be expanded as $$g(x)=\sum_{n=-\infty}^\infty c_n e^{i\pi n x} \,.$$ The eveness requires $$\frac{e^{-i\pi x/2}}{2} \, g(-x) = \frac{e^{i\pi x/2}}{2} \, g(x)$$ which after some algebra gives $$c_n = c_{-n-1} \quad n=0,1,2,...\,.$$ Finally $$\frac{e^{i\pi x/2}}{2} \, g(x)= \frac12 \sum_{n=-\infty}^\infty c_n e^{i\pi x(n+1/2)}= \frac12 \left( \sum_{n=-\infty}^{-1} + \sum_{n=0}^\infty \right) c_n e^{i\pi x(n+1/2)} \\ =\frac12 \sum_{n=0}^\infty c_{-n-1} e^{-i\pi x(n+1/2)} + \frac12 \sum_{n=0}^\infty c_n e^{i\pi x(n+1/2)} = \sum_{n=0}^\infty c_n \cos(\pi x (n+1/2)) \, .$$