# Tangent of angles approaching 90 degrees

Last week a colleague came to me with a puzzle. He asked me to punch in the tangent of 89 degrees into a nearby TI-83 calculator.

$\tan(89^\circ) = 57.28996163$

He asked me what was surprising about this number. I wasn’t surprised. I didn’t have an answer for him, although in retrospect I probably should have. He had to tell me this was the number of degrees in a radian. Oh! So it is.

Even further, he said, try punching in tan(89.9), or tan(89.99), etc.

$\begin{array}{rl} \tan(89^\circ) &= 57.28996163\\ \tan(89.9^\circ)&=572.9572134\\ \tan(89.99^\circ) &= 5729.577893\\ \tan(89.999^\circ) &= 57295.77951\end{array}$

Each one is (about) ten times the previous. (With the TI-83, replacing the last result with each new one, I didn’t see the “about” until later.) This is kinda neat! His question to me was: WHY is this true?

Tangent is a function that accepts an angle and spits out a ratio of lengths. It seems weird that the answer for tan(89) looks like a number of degrees. It is a unitless output, though, and ~57 degrees per radian is also unitless, so I suppose this isn’t much of an issue. The question is, why does it appear that

$\tan(89^\circ) = \cfrac{180^\circ}{\pi \text{ rad}}\text{ ?}$

Ditching the degree measure,

$\tan\left(\cfrac{\pi}{2}-\cfrac{\pi}{180}\right) \stackrel{?}{=} \cfrac{180}{\pi}$

My problem was in trying to first tackle this question visually.

An equivalent image:

I tried to explain this by imagining rolling the circle over the tangent line, wrapping the line around the circle, etc. I didn’t get anywhere.

I also tried considering the fact that there’s nothing particularly special about degree measure, except for the fact that 360 is an easy to divide number. Does this happen with other angle units? For example, what about a unit that was, instead of 1/360 of a circle, a larger 1/100 of a circle? We could instead take the equation above and ask,

$\tan\left(\cfrac{\pi}{2}-\cfrac{\pi}{100}\right) \stackrel{?}{=} \cfrac{100}{\pi},$

Is the tangent of one one-hundredth of a circle short of $\pi/2$ equal to the number of hundredths of a circle in a single radian? It looks to be true!

$\begin{array}{rl} \tan\left(\cfrac{\pi}{2}-\cfrac{\pi}{100}\right)&= 31.82051595\dots\\\\ \cfrac{100}{\pi} &=31.83098862\dots \end{array}$

But this is only approximate. We could extend this to any fractional unit $1/n$ of a circle:

$\tan\left(\cfrac{\pi}{2}-\cfrac{2\pi}{n}\right) \approx \cfrac{n}{\pi}$

Using this different unit, where the approximation is less accurate, I was able to see that the degree version wasn’t exactly true, either. It definitely looks like dividing the circle into a larger number (360, rather than 100) yields a closer approximation:

https://www.desmos.com/calculator/vqbu4zni5p?embed

I was comfortable in concluding now that this wasn’t just a coincidence that relied on degree measure, and could extend this to include using 89.9, 89.99 etc degrees as well. In fact, tacking on .9s to the 1/nths of a circle units works. Just plugging in a bunch of numbers, it looks like

$\tan\left(\cfrac{\pi}{2}-\cfrac{2\pi}{n}\ 10^{-a}\right) \approx \cfrac{n}{\pi}\ 10^a$

works for any n, and also extends to any power a, not just the integers.

https://www.desmos.com/calculator/ctcicykylj?embed

The question remained, why is this true? Now that I saw it’s only an approximation, I realized that I should be going about this algebraically from the start.

A trick, called the small angle approximation, is used in physics often to get rid of pesky sines and tangents when you’d rather just have an expression with the angle inside.

$\begin{array}{rl} \sin x &\approx x\\ \tan x& \approx x \qquad\text{when }x\ll1 \end{array}$

This behavior is clear when the functions are written in their Taylor series form:

$\begin{array}{rl} \sin x &= x - \cfrac{x^3}{6} + \cfrac{x^5}{120} - \cfrac{x^7}{5040} +\dots \\\\ \tan x &= x +\cfrac{x^3}{3} + \cfrac{2x^5}{15} +\cfrac{17x^7}{315}+\dots \end{array}$

When x is real small, all the higher power terms get super small, and the approximation becomes more accurate.

This approximation was my first thought, but there’s a problem: it works for small angles, but my colleague’s puzzle was about angles near 90 degrees. In fact, we can’t even fudge the Taylor series of tangent near here, because there is no Taylor series around 90 degrees. (This is a consequence of the fact that tan(x) blows up to infinity at 90 degrees.)

The problem is solved by noting that working with tangent near 90 degrees is the same as working with another trig function, cotangent, near 0 degrees.

$\tan\left(\cfrac{\pi}{2}-x\right) = \cot(x) = \cfrac{1}{\tan(x)}.$

Setting everything up:

$\begin{array}{rl} \tan\left(\cfrac{\pi}{2}-\cfrac{2\pi}{n}\ 10^{-a}\right) &= \left(\tan\left(\cfrac{2\pi}{n}\ 10^{-a} \right) \right)^{-1} \\ & = \left(\left(\cfrac{2\pi}{n}\ 10^{-a} \right) +\cfrac{1}{3}\left(\cfrac{2\pi}{n}\ 10^{-a} \right)^3 +\dots \right)^{-1}\\ &\approx \left(\cfrac{2\pi}{n}\ 10^{-a} \right)^{-1} \qquad \text{(the approximation)}\\ &= \cfrac{n}{2\pi}\ 10^a \end{array}$

Done! Having the $10^a$ instead of any old number is unneccessary — this works for any multiple. However, integer a‘s makes the trick of having the same digits show up in tan(89), tan(89.9), etc. work.

So, we can show this algebraically. I just wish I had a nice geometric argument.