Sine Infinite Product

Summary: §158: $\sin z=z{\prod }_{k=1}^{\infty }(1-z^2/k^2{\pi }^2)$. Euler’s celebrated infinite-product representation of the sine, obtained by substituting $x=iz$ in the hyperbolic-sine product.

Sources: chapter9

Last updated: 2026-04-29

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Statement

$\boxed{\sin z=z\prod _{k=1}^{\infty }\left(1-\frac{z^2}{k^2{\pi }^2}\right)=z\left(1-\frac{z^2}{{\pi }^2}\right)\left(1-\frac{z^2}{4{\pi }^2}\right)\left(1-\frac{z^2}{9{\pi }^2}\right)\cdots }$

Equivalently, factoring each $1-z^2/k^2{\pi }^2=(1-z/k\pi )(1+z/k\pi )$,

$\sin z=z\left(1-\frac{z}{\pi }\right)\left(1+\frac{z}{\pi }\right)\left(1-\frac{z}{2\pi }\right)\left(1+\frac{z}{2\pi }\right)\left(1-\frac{z}{3\pi }\right)\left(1+\frac{z}{3\pi }\right)\cdots$

(source: chapter9, §158).

Derivation

By eulers-formula, $\sin z=(e^{iz}-e^{-iz})/(2i)$. Set $x=iz$ in the §156 formula

$\frac{e^x-e^{-x}}{2}=x{\prod }_{k=1}^{\infty }\left(1+\frac{x^2}{k^2{\pi }^2}\right).$

The left-hand side becomes $(e^{iz}-e^{-iz})/2=i\sin z$, and $x=iz$ gives leading factor $iz$ and replaces $x^2=-z^2$ in each product term. Dividing both sides by $i$:

$\sin z=z{\prod }_{k=1}^{\infty }\left(1-\frac{z^2}{k^2{\pi }^2}\right).$

How the zeros encode the function

$\sin z=0$ iff $z=k\pi$ for some integer $k$ (positive, negative, or zero). The product exhibits this directly: the leading $z$ vanishes at $z=0$, and the factor $(1-z^2/k^2{\pi }^2)$ vanishes at $z=\pm k\pi$. Every zero of $\sin z$ is accounted for, with the right multiplicity.

Compare the power series

$\sin z=z-\frac{z^3}{3!}+\frac{z^5}{5!}-\cdots$

The series side encodes the behavior near $0$; the product side encodes the zeros everywhere on $\mathbb{R}$. Euler’s analytical philosophy — every transcendental function is the limit of a polynomial — makes both representations available.

Vanishing condition

“Whenever the arc has a length such that any of the factors vanishes, that is when $z=0$, $\pm \pi$, $\pm 2\pi$, etc., or generally when $z=\pm k\pi$, where $k$ is any integer, then the sine of that arc must equal zero. But this is so obvious, that we might have found the factors from this fact” (source: chapter9, §158). Reading backwards: knowing the zeros suffices to write down the product.

Why this matters

This is the formula that underlies Euler’s solution of the Basel problem: ${\sum }_{k=1}^{\infty }1/k^2={\pi }^2/6$. Sketch: equate the coefficient of $z^3$ in the power series $\sin z=z-z^3/6+\cdots$ with the coefficient of $z^3$ obtained by expanding the product. From the product, the $z^3$ coefficient is $-z\sum 1/k^2{\pi }^2$, hence

$-\frac{1}{3!}=-{\sum }_{k=1}^{\infty }\frac{1}{k^2{\pi }^2}⟹{\sum }_{k=1}^{\infty }\frac{1}{k^2}=\frac{{\pi }^2}{6}.$

Higher-order sums $\sum 1/k^4={\pi }^4/90$, etc., follow from comparing higher coefficients. Euler develops these in Chapter 10 of the Introductio via Newton’s identities (see basel-problem and zeta-at-even-integers).

The same idea (compare power-series coefficients with elementary symmetric polynomials of the reciprocal roots) generalizes to $\cos z$ via cosine-infinite-product, and to many further functions whose zero set is known.

Modern footnote

Euler’s derivation manipulates infinite products as if they were finite, with no convergence checks. The formula nonetheless turns out to be true, and the rigorous version is a special case of the Weierstrass factorization theorem for entire functions of order $1$. Euler’s instinct that “function = polynomial whose zeros we list” is the origin of that whole branch of complex analysis.

Related pages

• cosine-infinite-product
• exponential-infinite-product
• trinomial-factor
• factorization-of-an-plus-minus-zn
• sine-and-cosine-series
• eulers-formula
• pi
• newtons-identities
• basel-problem
• zeta-at-even-integers
• cotangent-partial-fraction
• linear-factors-of-sine-cosine
• wallis-product
• chapter-9-on-trinomial-factors
• chapter-10-on-the-use-of-the-discovered-factors-to-sum-infinite-series
• chapter-11-on-other-infinite-expressions-for-arcs-and-sines