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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.

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    $\begingroup$ Have you tried the double angle formula: $\sin(A) - \sin(B) = 2 \sin{\frac{(A - B)}{2} \cos\frac{(A + B)}{2}$? $\endgroup$
    – Peter K. ♦
    May 10, 2021 at 14:05
  • $\begingroup$ This is a good example of aliasing. Assuming n is an integer than $x[n]$ is a time discrete signal with a sample rate of 1Hz and a Nyquist Frequency of 0.5Hz. Your second term has a frequency of 1/5 Hz which is fine. However the first term has a frequency of 6/5 Hz. It's above Nyquist and aliases down to 1/5Hz and cancels the second term. $\endgroup$
    – Hilmar
    May 10, 2021 at 15:01
  • $\begingroup$ @Hilmar: I get most of what you said but did you make up the sample rate of 1HZ or figure it out from something in the OP's question. Thanks. $\endgroup$
    – mark leeds
    May 11, 2021 at 1:00
  • $\begingroup$ @markleeds: A sine wave with frequency $f$ is given by $\sin(2\pi ft)$. If you sample it at $t=nT=n/f_s$ you get $\sin(2\pi n f/f_s)$. So only the ratio $f/f_s$ is given, no absolute values. You could as well assume a sampling frequency of $\pi^2$ GHz if you like. $\endgroup$
    – Matt L.
    May 11, 2021 at 10:32
  • $\begingroup$ got it. I got confused because I thought the 1HZ was derived. Hilmar set $f_s = $ 1HZ so that $f/f_{s}$ is greater than $1/2$ for the first sine wave so that there is aliasing. thanks. $\endgroup$
    – mark leeds
    May 11, 2021 at 18:42

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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.

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