sine-and-cosine-series

Power Series for Sine and Cosine

Summary: §134 of Chapter 8. Letting $z$ be infinitely small (so $\sin z=z$, $\cos z=1$) and $n$ infinitely large with $nz=v$ a finite arc, the De Moivre expansions of $\cos nz$ and $\sin nz$ collapse — by the same $(j-m)/(nj)=1/n$ identity used in Chapter 7 — into the canonical power series

$\cos v=1-\frac{v^2}{2!}+\frac{v^4}{4!}-\frac{v^6}{6!}+\cdots ,\sin v=v-\frac{v^3}{3!}+\frac{v^5}{5!}-\frac{v^7}{7!}+\cdots .$

Euler immediately tabulates $\sin (m\pi /(2n))$ and $\cos (m\pi /(2n))$ as series in $m/n$ with stunning 28-digit numerical coefficients.

Sources: chapter8 (§134)

Last updated: 2026-04-27

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The infinitesimal/infinite move

The §133 De Moivre expansions

$\cos nz=(\cos z{)}^n-(\genfrac{}{}{0ex}{}{n}{2})(\cos z{)}^{n-2}(\sin z{)}^2+(\genfrac{}{}{0ex}{}{n}{4})(\cos z{)}^{n-4}(\sin z{)}^4-\cdots ,$

$\sin nz=n(\cos z{)}^{n-1}\sin z-(\genfrac{}{}{0ex}{}{n}{3})(\cos z{)}^{n-3}(\sin z{)}^3+(\genfrac{}{}{0ex}{}{n}{5})(\cos z{)}^{n-5}(\sin z{)}^5-\cdots$

hold for every positive integer $n$ and every arc $z$. Euler now does the move that defines the Introductio: take $z$ to be infinitely small, so

$\sin z=z,\cos z=1,$

(an infinitesimal-arc identity Euler accepts as obvious) and take $n$ infinitely large, with $nz$ equal to a finite arc $v$. Set $z=v/n$.

Under these substitutions:

• $(\cos z{)}^{n-k}=1^{n-k}=1$ for every $k$.
• $(\sin z{)}^k=z^k=v^k/n^k$.
• The binomial coefficient $\left(\genfrac{}{}{0ex}{}{n}{k}\right)=\frac{n(n-1)(n-2)\cdots (n-k+1)}{k!}$ has $k$ factors in the numerator. With $n$ infinite, each factor $(n-j)/n=1$ for finite $j$, so $n(n-1)\cdots (n-k+1)=n^k$, and $\left(\genfrac{}{}{0ex}{}{n}{k}\right)\cdot 1\cdot v^k/n^k=v^k/k!$.

This is the same coefficient-collapse used to derive $e^z=\sum z^n/n!$ in §116 from $(1+z/j{)}^j$ with $j$ infinite.

The cosine series

Substituting into §133’s $\cos nz$ expansion:

$\cos v=1-\frac{v^2}{2!}+\frac{v^4}{4!}-\frac{v^6}{6!}+\frac{v^8}{8!}-\cdots .$

(source: chapter8, §134). The series is alternating because the §133 expansion alternates in sign on $\left(\genfrac{}{}{0ex}{}{n}{2k}\right)(\sin z{)}^{2k}$.

The sine series

Substituting into §133’s $\sin nz$ expansion:

$\sin v=v-\frac{v^3}{3!}+\frac{v^5}{5!}-\frac{v^7}{7!}+\frac{v^9}{9!}-\cdots .$

(source: chapter8, §134). Both series have the form ${\sum }_{k=0}^{\infty }(-1{)}^k{v}^{2k}/(2k)!$ and ${\sum }_{k=0}^{\infty }(-1{)}^k{v}^{2k+1}/(2k+1)!$. They converge for all real $v$ — and in fact for all complex $v$, though Euler does not press this point.

Sample numerics: $\sin (m\pi /(2n))$ and $\cos (m\pi /(2n))$

Setting $v=(m/n)(\pi /2)$ — that is, taking $v$ to be the same fraction $m/n$ of the quarter-arc — Euler substitutes $\pi /2=1.5707963267948966\dots$ and writes the resulting numerical series in powers of $m/n$. The first few coefficients (source: chapter8, §134):

$\sin \frac{m\pi }{2n}=\frac{m}{n}\cdot 1.5707963267948966192313216916-\frac{m^3}{n^3}\cdot 0.6459640975062462536557565838+\frac{m^5}{n^5}\cdot 0.0796926262461670451205055488-\cdots ,$

$\cos \frac{m\pi }{2n}=1-\frac{m^2}{n^2}\cdot 1.2337005501361698273543113745+\frac{m^4}{n^4}\cdot 0.2536695079010480136385833859-\cdots .$

The leading coefficient of $\sin (m\pi /(2n))$ is $\pi /2$ to 28 digits; the second is $(\pi /2{)}^3/3!=0.6459640975\dots$; etc. Euler tabulates 30 coefficients in each series. Since the formulas suffice for all sines and cosines once $m/n<1/2$ (i.e., arcs up to $\pi /4$ = 45°, the rest reachable by §128 reflections), and since powers of a fraction $<1/2$ shrink fast, “a few terms should be sufficient, especially if the number of decimal places is not so large” (source: chapter8, §134).

Tangent and cotangent

§135 reads off the tangent and cotangent series by long division of the §134 series:

$\tan v=\frac{v-v^3/3!+v^5/5!-\cdots }{1-v^2/2!+v^4/4!-\cdots },\cot v=\frac{1-v^2/2!+v^4/4!-\cdots }{v-v^3/3!+v^5/5!-\cdots }.$

Euler quotes 25-digit numerical series for $\tan (m\pi /(2n))$ and $\cot (m\pi /(2n))$ but defers the closed-form derivation to §197 of a later chapter. The first numerical formula reads

$\tan \frac{m\pi }{2n}=\frac{2mn}{n^2-m^2}\cdot 0.6366197723675+\frac{m}{n}\cdot 0.2975567820597+\frac{m^3}{n^3}\cdot 0.0186886502773+\cdots ,$

with $0.6366\dots =2/\pi$.

Why this derivation, and not Taylor series?

Calculus had not yet provided Taylor’s theorem in the form Euler would later use, and even after Taylor (1715) the convention in the Introductio was to derive series by algebraic manipulation rather than by differentiation. Euler’s argument is purely algebraic:

1. Identity for every integer $n$: §133 De Moivre.
2. Pass to the limit by treating $n$ as infinite and $z$ as infinitesimal: collapse the binomial coefficients.
3. Read off the resulting power series.

The same three-step pattern works for $e^z$ (Chapter 7), for $\log (1+x)$ (Chapter 7), for $\sin v$ and $\cos v$ (here), and for $\arctan t$ (§140). The Introductio’s analytic engine is this single trick repeated.

Related pages

• de-moivre-formula
• exponential-series
• infinitesimal-and-infinite-numbers
• binomial-series
• sine-and-cosine
• eulers-formula
• arctangent-series
• chapter-8-on-transcendental-quantities-which-arise-from-the-circle
• chapter-7-on-exponentials-and-logarithms-expressed-through-series