# Weird tangent property

• B
I'm sure that I am not the first one to notice this, but I found that for angles between 0 and 90 degrees, tan(90-10^n) approximately equals 5.7296*10^(-n+1). Is that purely a coincidence?

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for n= 0 it does not work. For n= 1 also it does not sorry this is not worth pursuing.

Well, that worked for n=1 approximately
in my calculater. Didn't work for n=0. Those two were the only positive values I could try. I can say without calculating that it won't work for negative values of n. Tried n=1.5 but, no. Can you tell that for which values is it working?

In general it cannot be true as most values of tan will be irrational and the right side of your hypothesis will give you a rational number for integral values of n!

Well, that worked for n=1 approximately
in my calculater. Didn't work for n=0. Those two were the only positive values I could try. I can say without calculating that it won't work for negative values of n. Tried n=1.5 but, no. Can you tell that for which values is it working?
It almost works for n=1, it works sort of for n=0 and works very well for any negative value of n. I made a mistake, I meant that it equals 5.7*10^(-n+1). Sorry about that.

In general it cannot be true as most values of tan will be irrational and the right side of your hypothesis will give you a rational number for integral values of n!
It's an approximation, of course they are irrational.

fresh_42
Mentor
It almost works for n=1, it works sort of for n=0 and works very well for any negative value of n. I made a mistake, I meant that it equals 5.7*10^(-n+1). Sorry about that.
Sure you didn't mean ##\tan(90° - 10^{-n}) \approx 5.7_3 \cdot 10^{n+1}\; \;(n \in \mathbb{N})\;##?

TeethWhitener
Gold Member
I'm guessing it has something to do with the Taylor series.
$$\tan(\frac{\pi}{2}-\theta) = \cot \theta$$
The Taylor series for cotangent is:
$$\cot \theta = \frac{1}{\theta} - \frac{1}{3}\theta - \frac{1}{45}\theta^3 + \cdots$$
I'm assuming you were working in degrees instead of radians, so we have to convert ##10^n## to radians:
$$\theta = \frac{10^n\pi}{180}$$
Plugging this back in:
$$\cot(\frac{10^n\pi}{180}) = \frac{180}{10^n\pi} - \frac{10^n\pi}{3\times 180} - \cdots$$
For ##n=1##, we have:
$$\cot(\frac{\pi}{18}) \approx \frac{18}{\pi} \approx 5.73$$
Because of the functional form of the series (with a ##1/\theta## term out front), as ##\theta## gets smaller, the first term in the series becomes a better and better approximation.

• Igael and fresh_42
It almost works for n=1, it works sort of for n=0 and works very well for any negative value of n. I made a mistake, I meant that it equals 5.7*10^(-n+1). Sorry about that.
I don't know if this is a co-incidence but I had noticed the same pattern a few days ago when I was plugging in 89, 89.9, 89.99, 89.999 etc in my calculater to notice the changes in values of tanx. And, every time the same digits appeared that you've written with the decimal point displaced. I ignored it and didn't take the trouble of making a formula. It might be a co-incidence that you're uploading this today.
I don't know how this is working. But, it's an approximation mostly for negative values of n and so, it works for a very small range. There are only two non-negative values values of n we can try.

Sure you didn't mean ##\tan(90° - 10^{-n}) \approx 5.7_3 \cdot 10^{n+1}\; \;(n \in \mathbb{N})\;##?
It's pretty much the same. But yours is a bit more rigorous, so I guess it's better. Although it also works for 80 degrees (n=-1 in your definition).

I'm guessing it has something to do with the Taylor series.
$$\tan(\frac{\pi}{2}-\theta) = \cot \theta$$
The Taylor series for cotangent is:
$$\cot \theta = \frac{1}{\theta} - \frac{1}{3}\theta - \frac{1}{45}\theta^3 + \cdots$$
I'm assuming you were working in degrees instead of radians, so we have to convert ##10^n## to radians:
$$\theta = \frac{10^n\pi}{180}$$
Plugging this back in:
$$\cot(\frac{10^n\pi}{180}) = \frac{180}{10^n\pi} - \frac{10^n\pi}{3\times 180} - \cdots$$
For ##n=1##, we have:
$$\cot(\frac{\pi}{18}) \approx \frac{18}{\pi} \approx 5.73$$
Because of the functional form of the series (with a ##1/\theta## term out front), as ##\theta## gets smaller, the first term in the series becomes a better and better approximation.
That's really interesting! So by decreasing n by one, we essentially multiply the first term by 10, and make the other ones smaller, so they get more insignificant. So the closer n gets to minus infinity, the better an approximation it is. That's good to know, stuff like that does not tend to be coincidental.