Why can $\tan$ have every real number as its value?

Question

If $\tan\theta=\frac{\sin\theta}{\cos\theta}$, and $\sin\theta,\cos\theta \in [-1,1]$, how can $f:(-\pi/2,\pi/2)\to\mathbb{R}:x\mapsto\tan(x)$ be surjective? How can we find for any $y\in\mathbb{R}$ an $x$ such that $\tan(x)=y$?

Answer 1

You seem to be asking if $x$ and $y$ are bounded how can their ratios be unbounded.

And the answer to that is that if $y$ is very small and close to $0$ then $\frac 1y$ is very large, and although $\frac xy$ is not quiet as large as $\frac 1y$, it can still be quite large.

So it shouldn't surprise us that if 1) $\tan \theta$ is continuous. and 2) $\tan \theta$ is (might be) unbounded (and could be both unbounded in the "positive" and "negative" directions-- an $\tan 0 = 0$) that we can conclude, by intermediate value theorem that A) $\tan$ is surjective.

But if you want something more direct:

Let $w$ be any real number.

Then $\sqrt{w^2 + 1} > \sqrt{w^2} = |w|$ so $|\frac {w}{\sqrt{w^2 + 1}}| < 1$.

So there exists a $\theta; -\frac \pi 2< \theta < \frac \pi 2$ so that $\sin \theta = \frac {w}{\sqrt{w^2 + 1}}$.

And $\cos \theta = \sqrt { 1 - (\frac {w}{\sqrt{w^2 + 1}}^2)} = \sqrt {1 - \frac {w^2}{w^2 + 1}} = \sqrt{\frac {w^2 + 1}{w^2 + 1} - \frac {w^2}{w^2 + 1}} = \sqrt {\frac 1{w^2 + 1}} = \frac 1{\sqrt{w^2 + 1}}$

There fore:

$\tan \theta = \frac {\sin \theta}{\cos \theta} = \frac {\frac w{\sqrt{w^2 + 1}}}{\frac 1{\sqrt{w^2 + 1}}} = w$.

And that's it.

So if I wanted to solve $\tan \theta = -9,827$, I would punch $\theta = \arcsin \frac {-9827}{\sqrt{(-9827^2 + 1)}} = \arcsin -0.99999999482240484906224130963566 = -89.99416955537648186081352217411^{\circ}$ into a calculator.

And $\tan -89.99416955537648186081352217411^{\circ} = -9827$.

Maybe a more reasonable problem is $\tan \theta = 7$. Punch in $\theta = \arcsin \frac{7}{\sqrt{50}} = 81.869897645844021296855612559093^\circ$. And $\tan 81.869897645844021296855612559093^\circ = 7$.

===== old, but complete calculations from scratch ====

Let $w \in \mathbb R$. We want to solve $w = \frac x{\sqrt{1 - x^2}}$ where $|x| < 1$.

So $\sqrt{1-x^2}w = x$

$(1-x^2)w^2= x^2$ (Note: squaring gives us an extraneous answer so we will get two rather than one solution. We will choose the one where $x$ and $w$ are the same sign.)

$(w^2+1)x^2 -w^2 = 0$

$x = \frac {0\pm \sqrt {0-4(-w^2)(w^2+1)}}{2(w^2 + 1)}= \frac {\pm 2|w|\sqrt{w^2+1}}{2(w^2+1)} = \frac w{\sqrt{w^2 + 1}}$

Note. If $|w|< 1$ then $\sqrt{w^2 + 1} > 1>|w|$ so $|x| = |\frac w{\sqrt{w^2 + 1}}| < 1$. And if $|w| \ge 1$ then $\sqrt{w^2 + 1} > \sqrt{w^2} = |w|$ so $|x| =|\frac w{\sqrt{w^2 + 1}}| < 1$.

So let $x = \frac w{\sqrt{w^2 + 1}}$ and let $y = \sqrt{1 - x^2} = \sqrt {1 - \frac {w^2}{w^2 + 1}} = \sqrt{\frac {w^2 + 1}{w^2 + 1} - \frac {w^2}{w^2 + 1}} = \sqrt {\frac 1{w^2 + 1}} = \frac 1{\sqrt{w^2 + 1}}$

Then $\frac xy = w$ and $|x| < 1$ and $|y|< 1$.

Let $\theta = \sin^{-1} x =\sin^{-1}\frac w{\sqrt{w^2 + 1}}$ and therefore $y = \sqrt{1 - \sin^2 \theta} = \cos \theta$.

And $\tan \theta = \frac {\sin \theta}{\cos \theta} = \frac xy = w$.

So $\tan$ is surjective.

Answer 2

Pick any number $m \in \Bbb R$ and draw a line of slope $m$ through the origin. Let $\theta$ be the angle from the $x$-axis to the line. Then $\tan\theta = m$.

Answer 3

$\cos \theta$ can get very close to $0$. When it does, $\sin \theta$ is very close to $\pm 1$. Do you have any problem with the fact that $\frac 1x$ can be very large when $x$ is small? Aside from the fact that $\tan \theta$ can be zero, it is the same thing. The statement you are asking why is it true is not true. $\tan \frac \pi 2 \not \in \Bbb R$, for example.

Answer 4

In any right angled triangle , hypotenuse is the longest side. see https://en.wikipedia.org/wiki/Hypotenuse

i.e, opposite side,adjacent side $\lt$ hypotenuse $$\sin\theta=\frac{\text{opposite side}}{\text{hypotenuse}}$$ and $$\cos\theta=\frac{\text{adjacent side}}{ \text{hypotenuse}}$$

both of them are proper fraction.

but$$\tan\theta=\frac{\text{opposite side}}{\text{adjacent side}}$$ is need not a proper fraction.

Answer 5

It is quite simple. First, you can agree that the tangent function is continuous. Next, it can be arbitrarily large, both positive and negative. That suffices to state that it takes all real outputs.

An even easier argument would be that the tangent function is the ratio between the legs of a right triangle. You can fix the side opposite to theta as 1 and the other as any real number. Hence, you are done.