A few days ago I was playing on my scientific calculator and I ran over an interesting little equation: $180\sin(1)$ is extremely close to $\pi$. At first I thought it was a coincidence, but then I tried $360\sin\left(\frac{1}{2}\right)$ and it was closer to pi. So then I tried out $90\sin(2)$ and it was farther from $\pi$. So I came up with the equation $y = 180x\sin\left(\frac{1}{x}\right)$, which I then simplified to $y = x\sin\left(\frac{180}{x}\right)$. Although it approaches $\pi$ slower (180 times slower), it is easier to understand what was going on on the desmos graphing calculator. It seems the larger the $x$ value, the closer it gets to $\pi$. I would like to know why exactly this happens.

EDIT: (Sine is in degrees not radians, just for clarification)

  • $\begingroup$ It appears that your calculator is taking its trig arguments in degrees. That is important in explaining this. $\endgroup$
    – 2'5 9'2
    Sep 14, 2014 at 2:09
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    $\begingroup$ Hint: $$\sin_\text{deg}(180/x) = \sin_\text{rad}(\pi/x).$$ $\endgroup$ Sep 14, 2014 at 2:09
  • $\begingroup$ Geometrically, consider the perimeter of a regular polygon with $x$ sides inscribed in the unit circle. $\endgroup$
    – user856
    Sep 14, 2014 at 2:14

4 Answers 4


Given that you are still using degrees for trig functions, I am not sure how familiar you are with calculus. But I will try to explain what's happening without calculus, and then show what happens with calculus.

As $x$ gets larger, it will cause your equation $y$ to increase. At the same time, the argument inside of $\sin$ is getting very small because you are dividing by $x$. When $\sin$ is evaluated near zero, the output will also be very near zero. You may have seen before that $$\sin(\theta) \approx \theta$$ when $\theta$ is small. Anyway, long story short, you are observing a mathematical battle where one part of the functions wants to diverge to infinity, and it is being multiplied by another part of the function that is tending toward zero. As you have noticed, the battle will balance out at $\pi$ in this particular case. Once you know (more) calculus, you will have tools to measure just how fast functions are growing or shrinking, and can evaluate limits like these. One useful tool is known as "L'Hospital's Rule" , which when applied to your function would say that $$lim_{x \rightarrow \infty} \left[x \cdot \sin\left(\frac{180}{x}\right)\right]$$ $$=lim_{x \rightarrow \infty} \left[\frac{ \sin\left(\frac{180}{x}\right)}{x^{-1}}\right]$$ $$=lim_{x \rightarrow \infty} \left[\frac{ \cos\left(\frac{180}{x}\right)\cdot \frac{-180}{x^2}}{-x^{-2}}\right]$$ $$=lim_{x \rightarrow \infty} \left[180\cos\left(\frac{180}{x}\right)\right]$$ Now you can argue that since $lim_{x \rightarrow \infty}\frac{180}{x}=0$ and $\cos(0)=1$ that $lim_{x \rightarrow \infty}\cos\left(\frac{180}{x}\right)=1$. So $$lim_{x \rightarrow \infty} \left[180\cos\left(\frac{180}{x}\right)\right]=180 \cdot 1=180$$ And since $180$ degrees is the same as $\pi$ radians, we have shown mathematically just what you have witnessed on your calculator.

EDIT: @Rodolvertice If you are interested in more limits like these, I will say that I think a very cool result comes from $$\lim_{x \rightarrow \infty} \left[\left(1+\frac{1}{x} \right)^x\right]$$

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    $\begingroup$ Thank you for explaining this in depth! (I have barely any knowledge of calc, 10th grade) Im sure I can figure out the rest now (what lim means, the hospital rules etc haha) $\endgroup$
    – vero
    Sep 14, 2014 at 3:06
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    $\begingroup$ +1 for meeting the OP where he is, rather than aiming over his head. $\endgroup$
    – mweiss
    Sep 14, 2014 at 3:18
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    $\begingroup$ @Rodolvertice A pleasure to do so :) It always makes me happy when people are naturally curious about the oddities of math. If you want to know more I highly recommend pursuing a math degree. You will learn things that blow your mind in ways you can't even imagine. And you will leave college with one of the most useful degrees available. (People may argue with me, but if you are good at math you can market yourself to a vast amount of careers.) $\endgroup$
    – graydad
    Sep 14, 2014 at 4:30


$$ \lim_{x \to \infty} x \sin{\pi/x} = \lim_{x \to \infty} \frac{\sin{\pi /x}}{1/x} = \ldots$$

L'Hôpital is your friend now.



$$\lim_{x \to \infty} x \sin_{deg} \frac{180}{x} = \lim_{x \to \infty} x \sin_{rad} \frac{\pi}{x} = \pi\lim_{y \to 0} \frac{\sin y}{y}$$where $y=\frac{\pi}{x}$


So I came to the same conclusion once, that as "x" approaches infinity, then the value "xSin(180/x)", in degrees, approaches pi. But I had a more deliberate approach at this the first time I figured it out.

It's quite simple... I treated every other regular polygon as we treat a circle... so that they too would have constants, as "pi" is to a circle... deriving a formula that would relate the number of sides of the polygon "n" and the longest distance from the center of the polygon "D" to the polygon's perimeter "C".Just as is "C = (pi)D".

I then ended up with "C or P = [nSin(180/n)]D". So that the constant of the shape "k" = "nSin(180/n)". I thought that, by logic, this SHOULD approach pi as "n" approaches infinity. It gave me this graph enter image description here

So you can see that when you calculate "xSin(180/x)", your value of "x" is actually the number of sides of that regular shape... and as "x" approaches infinity, i.e. infinite sides of circle, you get (pi).

You could actually try and play around to get to the same conclusion I did. It's pretty correct. I hope this helps :)

  • $\begingroup$ Welcome to math.se! Please read how to format mathematics and edit your post. $\endgroup$
    – Smylic
    Apr 4, 2017 at 14:22

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