Cosine Infinite Product

Summary: §158: $\cos z={\prod }_{k=0}^{\infty }(1-4z^2/(2k+1{)}^2{\pi }^2)$. The infinite-product representation of cosine, dual to the sine product and obtained by substituting $x=iz$ in the hyperbolic-cosine product.

Sources: chapter9

Last updated: 2026-04-29

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Statement

$\boxed{\cos z=\prod _{k=0}^{\infty }\left(1-\frac{4z^2}{(2k+1{)}^2{\pi }^2}\right)=\left(1-\frac{4z^2}{{\pi }^2}\right)\left(1-\frac{4z^2}{9{\pi }^2}\right)\left(1-\frac{4z^2}{25{\pi }^2}\right)\cdots }$

Equivalently, splitting each quadratic factor into linear factors,

$\cos z=\left(1-\frac{2z}{\pi }\right)\left(1+\frac{2z}{\pi }\right)\left(1-\frac{2z}{3\pi }\right)\left(1+\frac{2z}{3\pi }\right)\left(1-\frac{2z}{5\pi }\right)\left(1+\frac{2z}{5\pi }\right)\cdots$

(source: chapter9, §158).

Derivation

By eulers-formula, $\cos z=(e^{iz}+e^{-iz})/2$. Set $x=iz$ in the §157 formula

$\frac{e^x+e^{-x}}{2}={\prod }_{k=0}^{\infty }\left(1+\frac{4x^2}{(2k+1{)}^2{\pi }^2}\right).$

The left-hand side becomes $(e^{iz}+e^{-iz})/2=\cos z$, and $x^2=-z^2$ in each factor.

Zeros of $\cos z$

$\cos z=0$ iff $z=\pm (2k+1)\pi /2$ for $k=0,1,2,\dots$. The product exhibits exactly these zeros: the factor $1-4z^2/(2k+1{)}^2{\pi }^2=0$ when $z=\pm (2k+1)\pi /2$. Euler:

From this it again becomes obvious that when $z=\pm (2k+1)\pi /2$, then $\cos z=0$, which is clear from the nature of the circle.

(source: chapter9, §158).

Compare with the power series

§134 gave $\cos z=1-z^2/2!+z^4/4!-\cdots$. Equating the coefficient of $z^2$ on both sides:

$-\frac{1}{2!}=-{\sum }_{k=0}^{\infty }\frac{4}{(2k+1{)}^2{\pi }^2}=-\frac{4}{{\pi }^2}{\sum }_{k=0}^{\infty }\frac{1}{(2k+1{)}^2}.$

Hence

${\sum }_{k=0}^{\infty }\frac{1}{(2k+1{)}^2}=\frac{{\pi }^2}{8},$

the sum of reciprocals of odd squares. Together with ${\sum }_{k=1}^{\infty }1/k^2={\pi }^2/6$ from the sine product this gives ${\sum }_{\text{even}}1/k^2={\pi }^2/24$, i.e. $\frac{1}{4}\sum 1/k^2$, an internal consistency check.

A unified picture

The two products together give

$\frac{\sin z}{z}={\prod }_{k=1}^{\infty }\left(1-\frac{z^2}{k^2{\pi }^2}\right),\cos z={\prod }_{k=0}^{\infty }\left(1-\frac{4z^2}{(2k+1{)}^2{\pi }^2}\right).$

Multiplying them and using $\sin 2z=2\sin z\cos z$:

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

So the right-hand product splits into the even-$k$ part (giving $\sin z/z$ at argument $z$, that is the $k=2m$ terms become $1-z^2/m^2{\pi }^2$) and the odd-$k$ part (giving $\cos z$). This is the duplication formula in product form.

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