Show that a sinusoid having a frequency larger than one corresponds to a sinusoid having a frequency less than one.

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I am studying electrical engineering for fun online. There is this one solution to a question on an online textbook that does not make any sense to me.

The question is:

Show that $\cos(2\pi fn)=\cos(2 \pi (f+1)n)$, which means that a sinusoid having a frequency larger than one corresponds to a sinusoid having a frequency less than one. From here

The given solution is:

As $\cos(\alpha+\beta)=\cos(\alpha)\cos(\beta)−\sin(\alpha)\sin(\beta)$, $\cos(2\pi (f+1)n)=\cos(2 \pi fn)\cos(2 \pi n)−\sin(2\pi fn)\sin(2\pi n)=\cos(2\pi fn)$.

Unless I'm misreading their solution, they're basically saying that $\cos(\alpha+\beta)=\cos(\alpha)\cos(\beta)−\sin(\alpha)\sin(\beta) = \cos(\alpha)$? Which is obviously NOT right.

So does $\cos(2 \pi fn)=\cos(2 \pi(f+1)n)$? And if so how do you prove it?

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@Rahul helped me answer my question.

When n is an integer, $\cos(2\pi n)$'s value would be 1 always. And $\sin(2πn)$'s would be 0 always. . . Which essentially means that the $+2πn$ term in $\cos(2π(f+1)n)$ is a $360 * n (integer) degrees$ phase shift! Meaning $\cos(2πfn)=\cos(2π(f+1)n)$. I get it now! Thank you