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How does $$A_c\mu\cos (2\pi f_m t) \cos (2\pi f_c t)$$ become $$\mu A_c\frac{\cos(2\pi f_mt-2\pi f_ct)-\cos(2\pi f_mt+2\pi f_ct)}{2}?$$ What is the formula used?

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This is incorrect; what is true is that $$A_c\mu\cos 2\pi f_m t \cos 2\pi f_c t=\mu A_c\frac{\cos(2\pi f_mt-2\pi f_ct)+\cos(2\pi f_mt+2\pi f_ct)}{2}$$ (note the plus sign on the right-hand side). This comes from the identity $$\cos A\cos B=\frac{\cos(A-B)+\cos(A+B)}{2},$$ with $A=2\pi f_mt$ and $B=2\pi f_ct$. To prove this identity, note that $$\cos(A+B)=\cos A\cos B-\sin A\sin B$$ and $$\cos(A-B)=\cos A \cos B+\sin A\sin B,$$ so the $\sin$ terms cancel out when you add them and you get $$\cos(A+B)+\cos(A-B)=2\cos A\cos B.$$

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