# On The Range Of The Tangent

I'm struggling to understand why the range of tangent function is equals to all real numbers. Because tangent = opposite/adjacent right? But the sum of the opposite^2 + adjacent ^2 must be equal to the hypotenuse^2. So doesn't that put a limit on the value of denominator (and the numerator) of the ratio/tangent, therefore putting a limit on the range of the function?

Also, I don't get how the tangent function is able to produce every single real number. Because doesn't the fact that opposite^2 + adjacent ^2 must be equal to the hypotenuse^2 prevent the tangent function from producing certain numbers? Because for any given opposite value, the adjacent value can't just be any integer. It's square must add up with the square of the opposite to equal hypotenuse squared. Doesn't that prevent the tangent function from producing certain numbers, because there are only so many combinations of opposite values and adjacent values that can occur?

Can you please explain to me why all possible real numbers can by formed by the tangent, even with these limits? Can you please explain why this is not the case as simply as possible, without using calculus (because I haven't learnt it)? I realize when you plot it, it will go up to infinity. But I want to understand why this is the case, without using calculus

• Dividing by small numbers like $0.0001$ gives large numbers. And $\cos$ (in the denominator) can take on arbitrarily small numbers. – M. Winter Jun 25 '18 at 9:09
• @M.Winter The thing is though, tangent = opposite/adjacent right? But the sum of the opposite^2 + adjacent ^2 must be equal to the hypotenuse^2. So doesn't that put a limit on the value of denominator (and the numerator) of the ratio/tangent, therefore preventing it from being all real numbers? The value is further limited by the fact that neither x nor y can be negative. Can you please explain why this is not the case as simply as possible, without using calculus? I realize when you plot it, it will go up to infinity. But I want to understand why this is the case, without using calculus. – Ethan Chan Jun 26 '18 at 3:31

I'll try to answer this question in terms of definitions of the trigonometric functions in which we keep everything on the unit circle. As an aside, I do feel compelled to point out that the question as currently edited is completely incompatible with a unit-circle construction of the functions, but based on your comments I am willing to believe that you did intend to use the unit circle and that you only used the word "integer" in your edited question by mistake.

In the unit-circle definitions of the trigonometric functions, if you do them correctly, the only reason to mention a "triangle" is to show how the unit circle definitions agree with the SOH-CAH-TOA triangle-based definitions when the angle is between zero and a right angle. If you had never seen the SOH-CAH-TOA definitions and instead had used only the unit-circle definitions, there would be no need for you ever to think about a triangle in relation to the definitions of these functions.

To use the unit circle, you construct an angle $\theta$ counterclockwise from the positive $x$ axis. The ray at that angle outward from the origin intersects the circumference of the circle at some point. Let $(x,y)$ be the coordinates of that intersection point. Then \begin{align} \cos\theta &= x, \\ \sin\theta &= y, \\ \tan\theta &= \frac yx. \\ \end{align}

Forget "opposite" and "adjacent"; all you need is $x$ and $y.$

Now suppose you want $\tan\theta = v,$ where $v$ is some arbitrary real number, possibly very large. Let's see how you can find the exact values of $x$ and $y$ you need in order to construct an angle $\theta$ such that $\tan\theta = v.$

If $(x,y)$ is a point on the unit circle, we know that $x^2 + y^2 = 1.$ From the definition of the tangent function, we know that if the ray at angle $\theta$ intersects the unit circle at $(x,y),$ then $\tan\theta = \frac yx.$ That is, $v = \frac yx.$ Now observe that $$v^2 + 1 = \frac {y^2}{x^2} + 1 = \frac{y^2 + x^2}{x^2} = \frac 1{x^2}. \tag T$$

In general, $x$ in this equation could be positive, negative, or zero, but let's just try non-negative values of $x$ for now. Solving Equation $(T)$ for $x,$ taking only the non-negative solution and discarding any negative solution, we get $$x = \frac 1{\sqrt{v^2 + 1}}.$$

So that's $x.$ And of course since $v = \frac yx,$ and we now know both $v$ and $x,$ we can solve for $y$: $$y = vx.$$

That's all we need! Set $x$ and $y$ to these values (computed from the given tangent value $v$), draw the ray from the origin through $(x,y),$ and you have constructed the angle whose tangent is $v.$

The secret to getting a large tangent is that even though we have restricted both $x$ and $y$ to be in the range $-1$ to $1,$ so neither of them can get very large, there is nothing to stop us from making $x$ very small. When $x$ is tiny, $y$ is near $1,$ and $\frac yx$ is a large number.

Perhaps a geometric interpretation would help.

You said that $$\tan\theta = \frac{y}{x}$$ In this interpretation, $x$ and $y$ are the coordinates of the point where the terminal side of an angle in standard position (vertex at the origin, initial side on the positive $x$-axis) intersects the unit circle, in which case $x = \cos\theta$ and $y = \sin\theta$, so $$\tan\theta = \frac{\sin\theta}{\cos\theta}$$

Now consider the following diagram.

We will show that the terminal side of an angle $\theta$ in standard position intersects the line $x = 1$ at the point $(1, \tan\theta)$.

Suppose that the terminal side of an angle in standard position intersects the line $x = 1$ at the point $(1, t)$ (for angles in the second and fourth quadrants, extend the terminal side of the angle backwards so that it intersects the line $x = 1$). Then, by similar triangles, $$\frac{|t|}{1} = \frac{|\sin\theta|}{|\cos\theta|} = |\tan\theta| \implies |t| = |\tan\theta|$$

Now, we must check signs.

• If $\theta$ is a first-quadrant angle, $y = \sin\theta > 0$ and $x = \cos\theta > 0$, so $$\tan\theta = \frac{\sin\theta}{\cos\theta} > 0 \implies |\tan\theta| = \tan\theta$$
Moreover, in the first quadrant, the $y$-coordinate of a point on the line $x = 1$ is positive, so $t > 0$. Thus, $$t = |t| = |\tan\theta| = \tan\theta$$
Thus, $t = \tan\theta$.
• If $\theta$ is a second-quadrant angle, $y = \sin\theta > 0$ and $x = \cos\theta < 0$, so $$\tan\theta = \frac{\sin\theta}{\cos\theta} < 0 \implies |\tan\theta| = -\tan\theta$$
If we extend the terminal side of a second-quadrant angle backwards, it intersects the line $x = 1$ in the fourth quadrant, where the $y$-coordinate is negative, so $t < 0$. Hence, $$t = -|t| = -|\tan\theta| = \tan\theta$$
• If $\theta$ is a third-quadrant angle, $y = \sin\theta < 0$ and $x = \cos\theta < 0$, so $$\tan\theta = \frac{\sin\theta}{\cos\theta} > 0 \implies |\tan\theta| = \tan\theta$$ If we extend the terminal side of a third-quadrant angle backwards, it intersects the line $x = 1$ in the first quadrant, where the $y$-coordinate of a point is positive, so $t > 0$. Hence, $$t = |t| = |\tan\theta| = \tan\theta$$
• If $\theta$ is a fourth-quadrant angle, $y = \sin\theta < 0$ and $x = \cos\theta > 0$, so $$\tan\theta = \frac{\sin\theta}{\cos\theta} < 0 \implies |\tan\theta| = -\tan\theta$$ In the fourth quadrant, the $y$-coordinate of a point on the line $x = 1$ is negative, so $t < 0$. Hence, $$t = -|t| = -|\tan\theta| = \tan\theta$$
• If the terminal side of angle $\theta$ lies on the $y$-axis, then $\cos\theta = 0$, so $\tan\theta$ is undefined. This makes sense in terms of our diagram since if the terminal side of the angle lies on the $y$-axis, it is parallel to the line $x = 1$, so the terminal side of the angle will not intersect the line $x = 1$.

Notice that with this interpretation of $\tan\theta$, it follows that $\tan(\pi + \theta) = \tan\theta$, so tangent is periodic. Moreover, since there is no value $p > 0$ smaller than $\pi$ such that $\tan(\theta + p) = \tan\theta$ for all $\theta$, tangent has period $\pi$.

Find the range of $f(\theta) = \tan\theta$.

Consider what happens as $\theta$ increases from $0$ to $\pi/2$. At $\theta = 0$, $\tan\theta = 0$ since the terminal side of the angle intersects the line $x = 1$ at the point $(1, 0)$. As $\theta$ increases, so does $\tan\theta$ since the point where the terminal side of the angle intersects the line $x = 1$ moves up the line. As $\theta$ approaches $\pi/2$, $\tan\theta$ increases toward $\infty$. Since $\tan\theta$ is continuous in the interval $[0, \frac{\pi}{2})$, it assumes all nonnegative real values.

Consider what happens as $\theta$ decreases from $0$ to $-\pi/2$. At $\theta = 0$, $\tan\theta = 0$. As $\theta$ decreases, so does $\tan\theta$ since the point where the terminal side of the angle intersects the line $x = 1$ moves down the line. As $\theta$ approaches $-\pi/2$, $\tan\theta$ decreases toward $-\infty$. Since $\tan\theta$ is continuous in the interval $(-\frac{\pi}{2}, 0]$, it assumes all nonpositive real values.

Thus, in the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$, $\tan\theta$ assumes all real values. Hence, $f(\theta) = \tan\theta$ has range $(-\infty, \infty) = \mathbb{R}$.

• Thanks, but I'm a little confused on 2 things. First, if we are using similar triangles, why are you dividing t/1 and sin(theta)/cos(theta)? I thought in similar triangles, you divide the equivalent side of the 2 triangles together, so shouldn't it be tan(theta)/sin(theta)=1/cos(theta)? Second, what do you mean extending the second quadrant angle backwards? I'm not sure what you mean by that. Can you draw it out, or just give me some way to visualize that please? – Ethan Chan Jun 28 '18 at 4:33
• Let $a, b, c, d$ be integers with $b, d \neq 0$. Recall that $$\frac{a}{b} = \frac{c}{d} \iff ad = bc$$ Thus, if $c, d \neq 0$, we have $$\frac{a}{c} = \frac{b}{d} \iff ad = bc$$ Consequently, if $b, c, d \neq 0$, then $$\frac{a}{b} = \frac{c}{d} \iff \frac{a}{c} = \frac{b}{d}$$ Substituting $\tan\theta$ for $a$, $\sin\theta$ for $b$, $1$ for $c$, and $\cos\theta$ for $d$ gives $$\frac{\tan\theta}{\sin\theta} = \frac{1}{\cos\theta} \iff \frac{\tan\theta}{1} = \frac{\sin\theta}{\cos\theta}$$ provided $\sin\theta \neq 0$. Notice that when $\tan\theta$ is defined, $\cos\theta \neq 0$. – N. F. Taussig Jun 28 '18 at 10:17
• If $\theta$ is a first- or fourth-quadrant angle, the ray representing the terminal side of the angle intersects the line $x = 1$ at the point $(1, \tan\theta)$. If $\theta$ is a second- or third-quadrant angle, the ray representing the terminal side of the angle intersects does not intersect the line $x = 1$. However, if that ray is extended backwards over the origin to form a line, that line will intersect the line $x = 1$. For a second- or third-quadrant angle $\theta$, we define the point where the line that includes the ray intersects the line $x = 1$ to be $\tan\theta$. – N. F. Taussig Jun 28 '18 at 10:46
• In the diagram above, the terminal side of the third-quadrant angle $\pi + \theta$ does not intersect the line $x = 1$. However, when that ray is extended backwards over the origin, the line that includes the ray representing the terminal side of the angle $\pi + \theta$ intersects the line $x = 1$ at the point $(1, \tan\theta)$. Therefore, $\tan(\pi + \theta) = \tan\theta$. Now, draw a ray representing the terminal side of a second-quadrant angle $\theta$. If you extend that ray into a line, it will intersect the line $x = 1$ in the fourth quadrant. – N. F. Taussig Jun 28 '18 at 10:49

You are probably overlooking that $$\{1/x:\ x\in(0,1)\} = (1,\infty)$$ holds.

... doesn't the fact that $opposite^2 + adjacent^2$ must be equal to the $hypotenuse^2$ prevent the tangent function from producing certain numbers? Because for any given opposite value, the adjacent value can't just be any integer.

Indeed if you limit yourself to Pythagorean triples--that is, you only consider right triangles with integer sides--then it certainly is not true that the tangent function gives you all real numbers. The tangent of such a triangle can only be a rational number (and not just any rational number, but a member of a specific subset of the rational numbers).

For example, there is no integer-sided right triangle with an angle whose tangent is $\sqrt2.$

Whoever said "the range of [the] tangent function is [equal] to all real numbers" clearly did not mean to consider only integer-sided triangles. You must consider triangles whose sides can be any length, for example legs of $1$ and $\sqrt2,$ or legs of $e$ and $\pi,$ or even legs whose lengths are real numbers so obscure that we have not yet come up with a way to describe those specific numbers.

When you allow this, you can use the techniques in other answers to create angles with arbitrarily large tangent values.

But even if you restrict yourself to look only at integer-sided triangles, what is the upper bound of the hypotenuse? It seems silly to set such a bound, because we have Pythagorean triples (even primitive Pythagorean triples) with arbitrarily long hypotenuses. Why should we throw them away?

You can have a right triangle with all integer sides, the hypotenuse and opposite side have absolutely enormous lengths, and the adjacent side is (relatively) short.

Here is a way to create an integer-sided triangle with as large a tangent as you want. It is well-known that if $p$ and $q$ are integers, with $p > q$, then $a = p^2 - q^2,$ $b = 2pq,$ and $c = p^2 + q^2$ are sides of a right triangle. (You can confirm this yourself: plug these expressions for $a,$ $b,$ and $c$ into the two sides of the Pythagorean formula and do a little algebra to show that the formula is true. We get some additional useful information if $p$ and $q$ are relatively prime, but for this exercise it does not matter.)

Now set $q = 1.$ This says our triangle has sides $p^2 - 1,$ $2p,$ and $p^2 + 1.$ The hypotenuse must be $p^2 + 1,$ and if we choose angle $\theta$ adjacent to the side $2p,$ we get $$\tan\theta = \frac{p^2 - 1}{2p} = \frac12 p - \frac{1}{2p} > \frac12p - 1.$$

So do you want an angle whose tangent is greater than $N$? Then just set $p = 2N + 1.$ The tangent will never be an exact integer, of course, but it can be larger than any arbitrary integer.

• I see, but why wouldn't there be an upper bound for x^2+y^2=h^2? The tangent is given by a unit circle. In a unit circle, h^2 must be equal to 1, which sets an upper bound. So for any value of x, there is only one possible value of y (as it must add up to 1). So under these conditions, how can it still be all real numbers? And anyways, you wrote "The tangent will never be an exact integer, of course, but it can be larger than any arbitrary integer." I'm not sure what that means. Does it mean that there are certain integers it can't form? Or does that tan(theta) can still output all real nums? – Ethan Chan Jun 28 '18 at 2:55
• @EthanChan In the unit circle there are no integer right triangles (unless you accept the degenerate triangles with sides $1, 0, 1$). So of course everything I wrote in this answer is irrelevant in that situation, because this answer is meant to address your concerns about integer right triangles. If we set aside the requirement for integer sides and fit the triangle in the unit circle, then you don't need $y$ to be large. You just need $x$ to be small. – David K Jun 28 '18 at 3:42
• I'm sorry, but I was talking about unit circle triangles (should have made it more specific). But anyways. How does the range of tangent for all real numbers hold true for unit circle triangles? Thanks. – Ethan Chan Jun 28 '18 at 3:44
• @EthanChan When were you talking about unit circle triangles? Was it when you wrote, "the adjacent value can't just be any integer"? Was it when you deleted the part of the question that said $x$ and $y$ range between $-1$ and $1$? Yes, I looked at the change history to try to guess what you possibly could be thinking, and I see that your original question was very different from the question I answered. Perhaps you should stop trying to puzzle out my answer and give some attention to N.F. Taussig's answer instead, since that answer deals with the unit circle. – David K Jun 28 '18 at 3:57

Note that if $\theta=\frac\pi3$, then $\sin\theta=\frac{\sqrt3}2$ and $\cos\theta=\frac12$. Therefore,$$\tan\theta=\frac{\frac{\sqrt3}2}{\frac12}=\sqrt3,$$in spite of the fact that $\frac{\sqrt3}2,\frac12\in[-1,1]$.

Besides, since $\tan$ is continuous and$$\lim_{\theta\to\pm\frac\pi2}\tan\theta=\pm\infty,$$it follows from the intermediate value theorem that the range of $\tan$ is indeeed $\mathbb R$.

Consider a right angled triangle that is much taller than it is wide - for example, a triangle with sides $101, 5100, 5101$. Note that we know this is a right angled triangle by applying Pythagoras's Theorem:

$101^2 + 5100^2 = 10201 + 26010000 = 26020201 = 5101^2$

So, besides the right angle, this triangle has two other angles, one that is very small - let's call this ang;e $\epsilon$ - and one that is just less than 90 degrees - this one is $90^o-\epsilon$.

Then

$\tan(\epsilon) = \frac{\text{opposite}}{\text{adjacent}} = 101/5100 \approx 0.0198$

and

$\tan(90^o-\epsilon) = \frac{\text{opposite}}{\text{adjacent}} = 5100/101 \approx 50.495$

By considering even thinner and taller triangles you can see that the tangent of very small angles can be as small as you like, whereas the tangent of angles close to 90 degrees can be as large as you like.

Your question seems to be based on a misunderstanding. You assume that $\tan x$ is defined as "opposite over adjacent" in a right angled triangle. This is not the definition.
The definition is that $\tan x = \frac{\sin x}{\cos x}$, where x is any angle, i.e. $x \in (-\infty, +\infty)$.
The function $\sin x$ can be positive (or negative) simultaneously with the $\cos x$ function being negative (or positive). If you agree, this does away with your objection that $\tan x$ can only be positive.
Your objection that the value of $\tan x$ is somehow limited because in a right triangle $x^2 + y^2 = h^2$, is difficult to understand. If I pick a random real number $x$ and a random real number $y$, their ratio is $\frac{y}{x}$. This ratio is unaffected by calculating the value of $h$.
• No, as you can just divide both sides by $h^2$ to get $(\frac{x}{h})^2 + (\frac{y}{h})^2= 1$. The ratio of the two sides is then $\frac{y}{h}/\frac{x}{h}= \frac{y}{x}$, i.e. the same as before. – Jens Jun 28 '18 at 16:18