Check if the system: $\sin \left(\frac{2\pi n x}{b-a} \right), \cos \left(\frac{2\pi n x}{b-a} \right)$is orthogonal over the interval [a=12, b=14].
My work thus far: $$\begin{align} \int^{14}_{12} \sin\left(\frac{2\pi n x}{14}\right) \cos\left(\frac{2\pi n x}{14}\right)dx &= \int^{14}_{12} \frac{1}{2}\sin\left(\frac{2\pi n x}{7}\right)dx \\[4pt] &= \frac{7}{4n}\pi \left[-\cos\left(\frac{2\pi n x}{7}\right) \right]^{14}_{12} \\[4pt] &= \frac{7}{4n}\pi \left[-\cos\left(4\pi n\right) +\cos\left(\frac{12\pi n}{7}\right) \right] \end{align}$$
$$\frac{7}{4n}\pi \left[\sin \left(\frac{40\pi n}{7} \right) \sin \left(\frac{-16\pi n}{7} \right)\right]^{14}_{12} = $$ I don't know how to proceed to show that it is not equal to zero. Is it as simple as just writing $4\pi n \neq \frac{12 \pi n}{7}$ or is there some catch?