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How can the sine function be derived/proven?

The definition for $\sin(x)$ is of course given as $\frac{\text{opposite}}{\text{hyoptenuse}}$ of a right-angled triangle, which solving for $x$ can be had from the Maclaurin series:

$$\sin(x) = \sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}x^{2n+1}$$

Which can be simplified to (using euler's formula, which is also derived from maclaurin series):

$$\sin(x) = \large \frac{e^{ix}-e^{-ix}}{2i}$$

But since the use of maclaurin series assumes the derivative of the sin function to be known, which requires knowing the function, the proof becomes circular. How to prove that the identity holds without resorting to this circularity?

More than likely the derivation of the formula is going to use the pythagorean theorem:

$$a^2+b^2=c^2$$

Where $a$ and $b$ are the sides of the right-angled triangle and $c$ is the hypotenuse. From this it can be derived that the ratio between circumference of a circle and it's radius is (denoted by $\pi$):

$$\large \pi=\sum_{n=0}^{\infty} \frac{4(-1)^{n}}{2n+1} = 3.1415...$$

Which can be used to evaluate an angle (in units of radians):

$$\theta = s/r$$

$s=$ arc length, $r =$ radius.

And the equation for characterizing a unit-circle, also derived from pythagorian theorem:

$$x^2+y^2 = 1$$

Brethlosze
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Dole
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  • The sine function is not proved, rather, it is defined as the ratio of the opposite side to the hypotenuse of a right triangle. – zz20s Jan 02 '16 at 11:36
  • Take a look at this post and the answers there. This may become interesting for you. :) – Hosein Rahnama Jan 02 '16 at 11:43
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    It depends on what the purpose of your definition is. If it is to be introduced for the first time as a function, without assuming any knowledge of calculus, you would define $\sin\theta$ as the $y$ coordinate of a point on the unit circlewhose radius makes angle $\theta$ with the positive $x$ axis, measured positively anticlockwise. – David Quinn Jan 02 '16 at 13:02
  • @zz20s yes, but how to prove that sine function equals the formula in the question, without resorting to circular logic. (In contrast, circle logic is warmly welcomed). – Dole Jan 02 '16 at 13:10
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    I think the answer is that the sine function was made up to solve triangles, not that it was made first, then found to be able to solve triangles. – Simply Beautiful Art Jan 02 '16 at 13:43
  • @SimpleArt That doesn't explain where the formula to solve the said triangle comes from. I would like a proof of the formula without self-referentiality. – Dole Jan 02 '16 at 13:49
  • Oh, good point. (+1) – Simply Beautiful Art Jan 02 '16 at 13:56
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    It's not a problem to find the MacLaurin series of a function if the function IS the MacLaurin series. – John Joy Jan 02 '16 at 15:24
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    Read the @JohnJoy comment carefully. There is no need to know the derivatives of $\sin x$ if you define it as $\sin x = \sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}x^{2n+1}$ – Hosein Rahnama Jan 02 '16 at 21:20
  • @JohnJoy But that definition can only be arrived to, if the derivative is known. At least using Maclaurin series that is. – Dole Jan 02 '16 at 21:23
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    @Dole if the MacLaurin series definition needed to have its derivatives known (presumably you mean non-power series derivatives), then the MacLaurin series definition wouldn't be a definition at all. We can define it as such, and later "discover" that it has geometric interpretations (e.g."Oh my, what a coincidence").. – John Joy Jan 03 '16 at 03:51
  • @JohnJoy Perhaps we are talking past each other, but I don't see how it's possible to derive the macularuin series definition before first noting that sin(x) = cos(x) and likewise for cos(x). Turns it can be done via other means (geometrically) than actually using derivative rules on the function. – Dole Jan 03 '16 at 10:54

2 Answers2

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You can geometrically prove that $\frac{d}{dx}\sin(x)=\cos(x)$ as seen here. You can similarly prove that $\frac{d}{dx}\cos(x)=-\sin(x)$, which gives rise to the differential equation representation of $\sin(x)$. This differential equation has all the info you need about $\sin(x)$ to get the other things.

Stella Biderman
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  • I do use the triangle definition, I just want to understand how to get from that definition to the specific formula (in any form), that gives sin(x) in terms of x. – Dole Jan 02 '16 at 20:03
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    Oh I see. I edited to answer that – Stella Biderman Jan 02 '16 at 20:07
  • Thanks. Does a purely geometric derivation exist, or does deriving the formula require calculus/maclaurin series? – Dole Jan 02 '16 at 21:25
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    I doubt a purely geometric proof exists, because $e^x$ doesn't have any nice geometric properties AFAIK. It's possible there's a proof that had a decomposition of a triangle in such a way as to give rise to the infinite series, but I've never seen such a thing. – Stella Biderman Jan 02 '16 at 21:26
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Both sin and cosine are mappings from the interval $[0,1]$ to $[0,Pi/2]$ -- we are mapping arc length of a unit circle ($\theta$) to the abscissa ($x$) and ordinate (y) where $x^2 + y^2=1, \sin(\theta)=y, \cos(\theta)=x$

Whichever method you use (e.g., formal calculus or method of exhaustion which is what Euclid used) -- you will need to assume that arc length of very small arcs is the same as the ordinate: $\lim_{\theta \to 0}(\sin(\theta) - \theta)=0$

Once you have assumed the above, then you can calculate arc length ($\theta$) using $\sin$: $\theta = \lim_{n\to \infty} 2n \sin(\frac{\theta}{2n})$

Harish Chandra Rajpoot
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Sebastian Wolf
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