# Why are these two functions equal?

These functions are equal. But I don't understand why.

$$a \leftrightarrow f(x) =|\cos(2\pi x)|^2$$

$$b \leftrightarrow f(x) = \dfrac{\cos(4\pi x)}{2} + 0.5$$

Which results both in this plot:

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They are the same because $$|\cos(x)|^2=\cos^2(x)=\frac{1+\cos(2x)}{2}=\frac{\cos(2x)}{2}+\frac{1}{2}.$$ (Note that the absolute value doesn't do anything, because of the square; i.e., $x^2=|x|^2$ for any $x$.) See here. This can be derived as follows:
For any $x$ and $y$, we have $$\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y).$$ Thus, $$\cos(2x)=\cos^2(x)-\sin^2(x)$$ $$\cos(2x)=\cos^2(x)-(1-\cos^2(x))$$ $$\cos(2x)=2\cos^2(x)-1$$ and therefore $$\cos^2(x)=\frac{1+\cos(2x)}{2}$$
Here is another, more direct derivation, using complex exponential definition of cosine: $$\cos^2(x)=\left(\frac{e^{ix}+e^{-ix}}{2}\right)^2=\frac{(e^{ix})^2+2(e^{ix})(e^{-ix})+(e^{-ix})^2}{4}=$$ $$\frac{e^{i(2x)}+e^{-i(2x)}+2}{4}=\frac{\cos(2x)+1}{2}$$