# Nyquist Sampling rate

We have $z(t)=\cos(100πt)\cos^2(500πt)$ .

Find Nyquist sampling rate.

Well, I know that $f_s=2f_{max}$. My main problem is that I am not entirely sure if I could re-write the above signal as: $z(t)=(2π50t)\cos^2(2π250t)$, and say that $f_{max}=250$, therefore $f_s=500$Hz. That $\cos^2(\cdot)$ troubles me a bit. Anyone could help? Thanks.

$$\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}$$
on the $\cos^2(\cdot)$ term. You'll still have a $\cos(\theta)\cos(\phi)$ term but you can use the product-to-sum formula
$$\cos(\theta)\cos(\phi) = \frac{\cos(\theta - \phi) + \cos(\theta + \phi)}{2}$$