I'm trying to show that $\sin \left(\frac{n\pi}{R}x\right)$ and $\cos \left(\frac{m\pi}{R}x\right)$ are orthogonal using the trigonometric identity that $2\cos{mx}\sin{nx} = \sin((m+n)x) + \sin((m-n)x)$ on $L^{2}([0, R])$.
I could conclude this is true of the region of interest were $2\pi$-periodic, but since $\sin \left(\frac{n\pi}{R}x\right)$ and $\cos \left(\frac{m\pi}{R}x\right)$ have period $2R$ and the interval of interest is half that length, how can I conclude that
$$\int_{0}^{R}\cos \left(\frac{m\pi}{R}x\right)\sin \left(\frac{n\pi}{R}x\right)dx = \frac{1}{2} \left( -\frac{R \cos{\frac{\pi (m+n)}{R}x}}{m+n} \big\vert_{0}^{R} - \frac{R\cos{\frac{ \pi (m-n)}{R}}}{m-n} \big\vert_{0}^{R} \right) = 0 $$
Summary: since both functions of interest have period $2R$ and are being integrated over a region of length $R$, how can the integral be shown to be zero?
Indeed, this is not true. For instance, taking $R=\pi$ for simplicity, consider $n=1$ and $m=0$. Then the integral is $\int_0^\pi \sin( x)\,dx$, which is the integral of a positive integrand and is definitely not 0 (in fact it is 2).