# Trying to get the infinite product for $\sin x$

I start with the fact that the zeros of $\sin x$ are $n\pi$, $n\in\mathbb{Z}$. Therefore, it should be possible to express it as an infinite product: $$\sin x = x (x-\pi)(x+\pi)(x-2\pi)(x+2\pi)\cdots$$ $$= x\prod_{n=1}^\infty (x^2 - n^2\pi^2)$$ From here I cannot factor out $\pi^2$ as there are infinite terms and $\pi^\infty$ doesn't look reasonable. If I take out $n^2$ then I have $$x(n!)^2 \prod_{n=1}^\infty (\frac{x^2}{n^2} - \pi^2)$$ But this isn't complete. Infact, this looks wrong (n is defined inside the product afterall). What further manipulations would enable me to get the correct product: $$\sin x = x\prod(1-\frac{x^2}{n^2\pi^2})$$

• At the second line, the product you write is not convergent. Put $f_n(x):=n^2\pi^2-x^2$. Then put $g_n(x):=\frac 1{n^2}f_n(x)$: since for all $x$ we have $\lim_{n\to\infty}g_n(x)=1$, we can hope that the product converges. – Davide Giraudo Aug 26 '11 at 19:48
• It's not true that there are "infinite terms" unless there are some terms, each of which is infinite. Rather, there are infinitely many terms. – Michael Hardy Aug 26 '11 at 19:59
• many complex analysis books have discussions of infinite products (including the weierestrass product product formula and usually the specific examples of the sine and gamma functions) – yoyo Aug 26 '11 at 20:47

The infinite product representation for the sine function is $$\sin(\pi x)=\pi x\prod_{1}^\infty \left(1-\frac{x^2}{n^2}\right).$$ So in the post, $\sin x$ should be replaced by $\sin(\pi x)$. Then the issue raised in the post disappears.
To prove the result, one needs quite a bit more function theory than the informal type of reasoning about zeros. Note that for example $e^z$ has no zeros, but $e^z$ is not equal to the empty product.
• The factor $\pi x$ in the expansion of $\sin \pi x$ can be seen as a consequence of $\displaystyle\underset{x\rightarrow 0}{\lim }\displaystyle\frac{\sin \pi x}{x}=\pi$. – Américo Tavares Aug 26 '11 at 20:19