linear-factors-of-sine-cosine

Linear Factors of Sine and Cosine at Rational Angles

Summary: §184: evaluating the §158 sine and cosine products at $z=m\pi /(2n)$ rationalizes every quadratic factor $1-m^2/(kn{)}^2$ into a pair of linear factors $(kn-m)/kn$ and $(kn+m)/kn$. The result is two linear-factor product expressions for each of $\sin (m\pi /2n)$ and $\cos (m\pi /2n)$ — one direct, one via the co-function identity. This is the engine behind the Wallis product, the other trig products, and the [[log-pi-via-products|log-$\pi$]] / log-sine computations of Chapter 11.

Sources: chapter11

Last updated: 2026-05-01

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The setup

Recall the §158 products:

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

Substitute $z=m\pi /n$ in the sine product and $z=m\pi /(2n)$ in the cosine product (the factor of two cancels the $4$):

$\sin \frac{m\pi }{n}=\frac{m\pi }{n}\left(1-\frac{m^2}{n^2}\right)\left(1-\frac{m^2}{4n^2}\right)\left(1-\frac{m^2}{9n^2}\right)\cdots$

$\cos \frac{m\pi }{n}=\left(1-\frac{4m^2}{n^2}\right)\left(1-\frac{4m^2}{9n^2}\right)\left(1-\frac{4m^2}{25n^2}\right)\cdots$

Substituting $2n$ for $n$ — i.e. starting from arc $m\pi /(2n)$ instead of $m\pi /n$ — gives expressions with denominators $4n^2,16n^2,36n^2,\dots$ in the sine, and $n^2,9n^2,25n^2,\dots$ in the cosine. Each numerator $4k^2{n}^2-m^2$ or $(2k+1{)}^2{n}^2-m^2$ is a difference of squares, so factors as $(2kn-m)(2kn+m)$ or $((2k+1)n-m)((2k+1)n+m)$.

The four products of §184

Cleaning up the bookkeeping yields the first pair (source: chapter11, §184):

$\boxed{\sin \frac{m\pi }{2n}=\frac{m\pi }{2n}\cdot \frac{2n-m}{2n}\cdot \frac{2n+m}{2n}\cdot \frac{4n-m}{4n}\cdot \frac{4n+m}{4n}\cdot \frac{6n-m}{6n}\cdot \frac{6n+m}{6n}\cdots }$

$\boxed{\cos \frac{m\pi }{2n}=\frac{n-m}{n}\cdot \frac{n+m}{n}\cdot \frac{3n-m}{3n}\cdot \frac{3n+m}{3n}\cdot \frac{5n-m}{5n}\cdot \frac{5n+m}{5n}\cdots }$

Each factor is a rational number (assuming $m,n$ are integers) close to $1$ for large index, and the convergence is geometric in $1/k^2$.

The co-function identity gives a second pair

Use $\sin ((n-m)\pi /2n)=\cos (m\pi /2n)$, valid because $(n-m)\pi /(2n)+m\pi /(2n)=\pi /2$ (sine and cosine are co-functions across the right angle, §128). Apply the sine boxed formula above with $m\mapsto n-m$ and read off the result for $\cos (m\pi /2n)$:

$\cos \frac{m\pi }{2n}=\frac{(n-m)\pi }{2n}\cdot \frac{n+m}{2n}\cdot \frac{3n-m}{2n}\cdot \frac{3n+m}{4n}\cdot \frac{5n-m}{4n}\cdot \frac{5n+m}{6n}\cdot \frac{7n-m}{6n}\cdots$

(source: chapter11, §184, after the substitution $m\mapsto n-m$.)

Symmetrically, $\cos ((n-m)\pi /2n)=\sin (m\pi /2n)$, so applying the cosine boxed formula with $m\mapsto n-m$ gives

$\sin \frac{m\pi }{2n}=\frac{m}{n}\cdot \frac{2n-m}{3n}\cdot \frac{2n+m}{3n}\cdot \frac{4n-m}{5n}\cdot \frac{4n+m}{5n}\cdot \frac{6n-m}{7n}\cdot \frac{6n+m}{7n}\cdots$

So each of $\sin (m\pi /2n)$ and $\cos (m\pi /2n)$ has two infinite product representations. The redundancy is the source of every result in the rest of Chapter 11.

Where the linear factors come from

The §158 quadratic factor $1-z^2/k^2{\pi }^2$ at $z=m\pi /(2n)$ becomes

$1-\frac{m^2}{4k^2{n}^2}=\frac{4k^2{n}^2-m^2}{4k^2{n}^2}=\frac{(2kn-m)(2kn+m)}{(2kn{)}^2}=\frac{2kn-m}{2kn}\cdot \frac{2kn+m}{2kn}.$

Each $1-z^2/k^2{\pi }^2$ contributes two adjacent factors to the resulting product. The same algebra works for the cosine quadratic at $z=m\pi /(2n)$, with $2k$ replaced by $2k+1$. The whole maneuver is just “rationalize the §158 products at rational angles” — but doing so doubles the number of factors and exhibits each factor as the simplest possible rational number, which is exactly the form needed for the upcoming applications.

What the four products are good for

Use	Reference
Comparing two expressions for $\cos (m\pi /2n)$ → Wallis product for $\pi /2$	§185
Quotient of expressions for $\sin$ and $\cos$ → infinite products for $\tan$, $\cot$, $\sec$, $\csc$	§186
Replacing $m$ with $k$ → product for the ratio $\sin (m\pi /2n)/\sin (k\pi /2n)$	§187
$\log$ of the products + transposition → [[log-pi-via-products|series for $\log \pi$]]	§188–§190
Same on sine/cosine → [[log-sine-via-products|series for $\log \sin$, $\log \cos$]]	§191–§196

Why the rationalization matters

The §158 products are valid for all complex $z$ but are awkward for numerical work because each factor $1-z^2/k^2{\pi }^2$ involves ${\pi }^2$ — and computing ${\pi }^2$ is the very problem one is trying to solve. Restricting to rational angles eliminates $\pi$ from every factor: each entry in the §184 products is a fraction like $(2n-m)/(2n)$ with integer numerator and denominator. Logarithms of such fractions can be looked up in standard tables, and that is the door Euler walks through in §188–§198 to compute logs of $\pi$ and of trig functions to twenty-plus decimal digits.

Related pages

• sine-infinite-product
• cosine-infinite-product
• wallis-product
• trig-infinite-products
• log-pi-via-products
• log-sine-via-products
• trigonometric-addition-formulas
• chapter-11-on-other-infinite-expressions-for-arcs-and-sines