Why is $x[n]=\sin(\frac{12\pi n}{5})-\sin(\frac{2\pi n}{5})=0$ I was trying to find the period of $\sin(\frac{12\pi n}{5})-\sin(\frac{2\pi n}{5})$, each of the sinusoids has a period of 5 however their difference has a period of 1. It turned out that they're identical in the discrete-time domain.
besides actually solving the equation for reals and concluding integer roots, is there anything that explains this in terms of periodicity?
Thank you.
HINT:
$$\sin\left(\frac{12\pi n}{5}\right)=\sin\left(\frac{2\pi n}{5}+\frac{10\pi n}{5}\right)=\sin\left(\frac{2\pi n}{5}+2\pi n\right)$$
Taking into account that $\sin(x)$ is $2\pi$-periodic should make the result obvious.