Sine function as an infinite sequence

1. Mar 21, 2013

Appleton

I have a vague understanding of how to derive the sine function from a Maclaurin Sequence however this isn't helping me figure out why:
(1 - $\frac{x^{2}}{4π^{2}}$) (1 - $\frac{x^{2}}{9π^{2}}$) (1 - $\frac{x^{2}}{16π^{2}}$)... = $\frac{π^{2}}{x(x+π)}$$\frac{sin x - sin π}{x - π}$
Any help would be appreciated

2. Mar 21, 2013

Staff: Mentor

Where does this equation come from?
It is probably possible to prove this identity (assuming it is true) in some way, but probably not with a simple Taylor expansion of sin(x).

And where is the point in subtracting sin(pi)?

3. Mar 21, 2013

HallsofIvy

Staff Emeritus
This is certainly NOT true. At x= 0, the left side is 1 and the right side is 0.

4. Mar 21, 2013

Staff: Mentor

The right side is undefined at x=0, but it has 1 as limit. That is fine.

5. Mar 21, 2013

jbunniii

If you multiply both sides by $x(x-\pi)(x+\pi)/\pi^2$, and note that $\sin(\pi) = 0$ on the right hand side, you will end up with the product formula for the sine function, due to Euler:
$$\sin(x) = x \prod_{n=1}^{\infty}\left(1 - \frac{ x^2}{\pi^2 n^2}\right)$$
He obtained this formula by rather brashly viewing the sine function as a "polynomial" with infinitely many roots, namely $0, \pm \pi, \pm 2\pi, \pm 3\pi, \ldots$. Accordingly, he "factored" it as follows:
$$\sin(x) = kx(x \pm \pi)(x \pm 2\pi)(x \pm 3\pi)\ldots$$
for some constant $k$. The constant must be chosen so that $\lim_{x \rightarrow 0} \frac{\sin(x)}{x} = 1$, which forces
$$\sin(x) = x\left(1 \pm \frac{x}{\pi}\right)\left(1 \pm \frac{x}{2\pi}\right)\left(1 \pm \frac{x}{3\pi}\right)\ldots$$
Each plus/minus factor can be simplified as follows using the rule $(a-b)(a+b) = a^2 - b^2$:
$$1 \pm \frac{x}{n\pi} = 1 - \frac{x^2}{n^2 \pi^2}$$
and the result follows.

Of course the above is completely nonrigorous. Euler had the extraordinary ability to turn invalid manipulations into valid results! The same result can be obtained rigorously by using Fourier series. See for example Courant and John, Introduction to Calculus and Analysis I, page 602.

6. Mar 22, 2013

Appleton

thanks!

7. Mar 22, 2013

HallsofIvy

Staff Emeritus
Sorry- I completely overlooked the "x" in the denominator on the right.