Nyquist sampling rate with real and imaginary aliasing
From rad/sec to Hz

Usually we work in the frequency domain with Hz. Let me first make translations:


Signal \$\sin(6t)\$ is a signal of pulsation of \$ 6 rad.s^{-1}\$, which means it has a frequency \$f = \frac{6}{2\pi} \approx0.955 Hz\$
Same translation for \$\sin(5t)\$, it's a \$\approx~0.795Hz\$


Time view

If you subtract them, you would see something like the blue chart below, which is an approximation of the signal for 94.25 (30 \$\pi\$) seconds

Frequency view

The fourier transform will simply show one dirac for the \$6 rad.s^{-1}\$ signal, and another one for \$5 rad.s^{-1}\$. See the fourier transform below. Real part is in green, and imaginary part is in black.



Nyquist frequency must be greater than twice the highest frequency of the signal, so it must be greater than \$2*6 = 12 rad.s^{-1}\$. If you choose \$ 10 rad.s^{-1}\$, you will face a aliasing issue.