According to page 2 of this document, $\tan\theta=(y/x) \nRightarrow \theta=\tan^{-1}(y/x)$.
Example question: Convert the point $(\sqrt{2},-\sqrt{2},4)$ from Cartesian to Cylindrical coordinates.
One part of solving this question initially boils down to finding all $\theta \in [0,2\pi]$ such that $\tan\theta=\frac{y}{x}=-\frac{\sqrt{2}}{\sqrt{2}}=-1.$
The statement in the title confuses me because in examples I've seen so far (from different sources), the workings will go something like:
$$\tan\theta=-1 \implies \theta = \tan^{-1}(-1)=-\pi/4 \in \left(-\pi/2,\pi/2\right).$$ Hence $\theta = -\pi/4 +k\pi$ for $k\in\mathbb{Z}.$ We therefore have that $\theta$ is $3\pi/4$ or $7\pi/4$ for $\theta \in [0,2\pi].$
This in itself is actually now confusing me because I always thought $f^{-1}(y)=x \iff f(x)=y$, i.e. in this case that $\tan\theta =-1 \iff \theta= \tan^{-1}(-1)$.
But for $\theta=\tan^{-1}(-1)$, we know $\theta$ can only be $-\pi/4$ if it is to be in the range of $\tan^{-1}$, and with $\tan\theta=-1$ there are infinitely many solutions, so how can this be an if and only if relationship? Well according to the document hyperlinked above, it's not even a one-way relationship (contradicting the workings in the above box) and I'm rather confused.
Any clarification would be appreciated.
$\endgroup$ 32 Answers
$\begingroup$In general, the iff is only true if $f(x)$ is one to one. So the only error is that
$$f^{-1}(y)=x \iff f(x)=y\iff f(x)\text{ is one to one}$$
but $\tan(x)$ is not one to one. To make it one to one, we usually apply domain/range restrictions.
Since the solutions to $\tan(\theta)=a$ are easily generalized, we often split it up over branches:
$$\tan^{-1}(a,k)=\tan^{-1}(a)+\pi k$$
$\endgroup$ 12 $\begingroup$First, you should use the mathematical notation arctan, not the hand-held computer notation $\tan^{-1}$.
Second, as $\tan$ is not a bijection, since it is a periodic function, what is defined is the inverse function of its restriction to a domain on which its restriction is bijective. The chosen interval is $\;\Bigl]-\dfrac\pi2,\dfrac\pi 2\Bigr[$. Thus the full definition of $\arctan$ is $$\theta=\arctan\frac yx\iff \tan\theta=\frac yx\enspace\textbf{and}\enspace\boxed{\theta\in\Bigl]-\dfrac\pi2,\dfrac\pi 2\Bigr[.} $$
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