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Question:

Find the inverse of $$f: [\pi,\frac{3\pi}{2}] \rightarrow R, f(x) = \frac{1}{2}\cos(2x)$$

My attempt:

$$x=\frac{1}{2}\cos(2y)$$ $$\cos(2y)=2x$$ $$2y=\arccos(2x)$$ $$f^{-1}(x)=\frac{1}{2}\arccos(2x)$$

But the solution is apparently:

$$x=\frac{1}{2}\cos(2y-2\pi)$$ $$\cos(2y-2\pi)=2x$$ $$2y-2\pi=\arccos(2x)$$ $$2y=\arccos(2x)+2\pi$$ $$f^{-1}(x)=\frac{1}{2}\arccos(2x)+\pi$$

It would be great if someone could explain to me why there is an extra $-2\pi$ in the argument in the solution. I feel like it could have something to do with the domain of the original cos function. Thank you in advance!

1 Answers1

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This is because $\pi \le y \le 3\pi/2$, so $2\pi \le 2y \le 3\pi$. The function $\arccos$ is defined as the inverse of $\cos$ on $[0,\pi]$, so the range of $\arccos$ is $[0,\pi]$, and $2y$ is not in this interval. To fix this, subtract $2\pi$ from $2y$ to get a number $2y - 2\pi= z \in [0,\pi]$, for which then the following is correct (where the first "$\iff$" is by $2\pi$-periodicity):

$$ \cos (2y) = t \iff \cos z = t \iff z = \arccos t$$