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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.

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You can use the power reduction formula

$$\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2} $$

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

$$\cos(\theta)\cos(\phi) = \frac{\cos(\theta - \phi) + \cos(\theta + \phi)}{2}$$

to convert it to a sum. That will give you a sum of cosines from which you can pick out the highest frequency.

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