# 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.

• Have you tried the double angle formula: $\sin(A) - \sin(B) = 2 \sin{\frac{(A - B)}{2} \cos\frac{(A + B)}{2}$?
– Peter K.
May 10, 2021 at 14:05
• 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. May 10, 2021 at 15:01
• @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. May 11, 2021 at 1:00
• @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. May 11, 2021 at 10:32
• 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. May 11, 2021 at 18:42

$$\sin\left(\frac{12\pi n}{5}\right)=\sin\left(\frac{2\pi n}{5}+\frac{10\pi n}{5}\right)=\sin\left(\frac{2\pi n}{5}+2\pi n\right)$$
Taking into account that $$\sin(x)$$ is $$2\pi$$-periodic should make the result obvious.