# Solving For Sin Using Pi

Solving for sin using pi
I was messing around with calculating pi by finding the perimeter of a many sided polygon, and dividing it by the diameter (Like the thing Archimedes did). The equation I found was n(sin(180/n))=pi, where n is the number of sides the polygon has. I was wondering if there was any way to reverse this equation, making it so that you can solve for sine using pi.

Here is what I tried:
(Note: This is all in degrees because I do not know radians that well)
This is what I started with n(sin(180/n))=pi
Divide both sides by n
(sin(180/n))=pi/n
Substitute a for 180/n to get (sin(a))=pi/n.
This means 180/n=a
Then I multiplied both sides of that by n to get 180=a*n
And divided both sides by a to get 180/a=n
Now I have solved that equation for n, so I can substitute it back into my original equation
for n to get (sin(a))=pi/(180/a), which can be
simplified to sin(a)=pi*a/180. What this says is that the sin(a) is equal to
pi*a/180, which definitely isn't true. One interesting thing about this
equation is that it is the equation to convert degrees into radians. Also, if you graph it, you will is very close to the sin wave until about 25.
(If you graph this equation, make sure you are using degrees and not radians)

After trying this out, I did some reasearch and found there is no easy way to calculate sine. However, I would still like to know what was wrong with the math I did to simplify this.

Thanks.

I am 13 and this is my first question I have posted, so please excuse any mistakes.

• Try using $\frac{\pi}{2}$ instead of $180$ – waterfalls May 3 '16 at 21:26
• $n \sin 180^o/n is$ not EQUAL to $\pi$ .Only approximately. – DanielWainfleet May 3 '16 at 21:39
• Sorry. I forgot to say that pi is the limit for that equation if n=infinity – user336769 May 3 '16 at 21:47
• It's only approximate for very large n. So sin a ~= pi x a/ 180 will only be true for very small a. In radians this is simply sin a ~= a and furthermore a~= 0. So it's not exactly not true but it's not really true either. – fleablood May 3 '16 at 21:50

It turns out that the "equation" $n \sin(180/n))=\pi$ is not true. But, if $n$ is a large number then it is approximately true. That is to say, it is still not true on the nose, but as $n$ gets larger and larger, the difference between the two sides $n \sin(180/n) - \pi$ gets closer and closer to zero.

So the equation you deduced, namely $\sin(a) = \pi * a / 180$ is also not true, but it is approximately true. That is, as $a$ (measured in degrees) gets closer and closer to zero, the difference between the two sides $\sin(a) - \pi * a / 180$ gets closer and closer to zero. So in fact for small numbers $a$, the approximation $$\sin(a) \approx \pi * a / 180$$ is actually pretty accurate, and the accuracy gets better and better for values of $a$ closer and closer to zero. Try it out on your calculator for $a=1^\circ$, then for fractions of a degree such as $a=.1^\circ$, $a=.01^\circ$, and so on.

• Thanks. I noticed that it was pretty close for small numbers. However, it doesn't oscillate like a sine wave. Also are you saying the reason that the equation doesnt work is that in the original equation n has to be infinity, so when it isnt in the other equation it doesnt work? Also how do you make equations like that? – user336769 May 3 '16 at 21:31
• By the way, although you write "there is no easy way to compute $\sin$", nonetheless this approximation is exactly how people compute $\sin$ as long as $a$ is close enough to zero. For larger values of $a$, when the approximation you found gets too inaccurate, there are better approximations available, called "Taylor polynomials", and these are exactly how your calculator computes $\sin$. – Lee Mosher May 3 '16 at 21:35
• Regarding how to make the Taylor polynomials, you'll need to learn calculus. But here are the first few Taylor polynomials for $\sin(x)$, with a general pattern for producing all of them. I will express them in radians (you can convert back to degrees if you like): $$x, \,\,\, x - \frac{x^3}{3!}, \,\,\, x - \frac{x^3}{3!} + \frac{x^5}{5!}, \,\,\, x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!}, \,\,\, ...$$ and so on. If you don't know "factorial" notation, for example $7!$ is shorthand for $7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$. – Lee Mosher May 3 '16 at 21:48
• Regarding your question that "$n$ has to be infinity", that doesn't really make sense. Infinity is not a number, and there's no getting around that. But what does make sense is the concept of limits, which is something else you learn in calculus. When people say things like "larger and larger" or "closer and closer", they're talking about limits. – Lee Mosher May 3 '16 at 21:50

Let's take the problem in parts.

First: sin(x) is a function, and its inverse is the arcsin function. See Inverse Trigonometric Functions for definitions and properties.

Second: 180 degrees = pi radians. The equality you wrote turns to be a limit:

lim [n -> oo] n * sin(pi/n) = pi

In other words: when n grows to infinity, n * sin(pi/n) approaches pi. See Limit for a formal definition of limit. Assuming a big enough n, this is a near equality:

n * sin(pi/n) = pi

sin(pi/n) = pi/n

pi/n = arcsin(pi/n)

No more simplification is needed.

Related: lim [x -> 0] (sin(x) / x) = 1. This can be proved, see Squeeze Theorem for a sketch.

We have $\sin \pi/4=\cos \pi /4=1/\sqrt 2.$ And $\sin \pi /6=1/2$ and $\cos \pi /6=\sqrt 3 /2.$

You can go from case $n$ to case $2 n$ with $|\sin x/2|=\sqrt {(1-\cos x)/2}$ and $|\cos x/2|=\sqrt {(1+\cos x)/2}.$

We have $n \sin \pi /n<\pi <n \tan \pi /n$ for $n>2.$ To go from case $n$ to case $2 n$ for the $\tan$ , we have $|\tan x/2|=|1-\cos x)/\sin x|=|\sin x|/(1+\cos x)$ when $\sin x\ne 0.$

There is also a simple geometric way to explicitly obtain $\sin \pi /5$ and $\cos \pi /5.$