Worked solution: Euler's proof via the sine product formula (1734–1735)

Step 2 of 6: Maclaurin series of sin x / x

In plain words

Dividing by $x$ is like peeling off the one obvious layer of the function so that what remains can be compared honestly with a product built from its other zeros; this new function equals $1$ at $x=0$ and its power series is just the sine series with each power of $x$ shifted down by one.

\frac{\sin x}{x}=1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+\cdots

Detailed analysis

Dividing the known power series for $\sin x$ by $x$ removes the trivial zero at $x=0$ and gives a function equal to $1$ there; its remaining zeros are exactly $x=\pm\pi,\pm2\pi,\pm3\pi,\dots$ , the nonzero solutions of $\sin x=0$ .

Terms in this step

Maclaurin series: The power series expansion of a function around $x=0$ , obtained from its derivatives at that point; it lets a function be treated as an "infinite polynomial" in $x$ .
Zero of a function: A point where the function's value is exactly $0$ ; for $\sin x$ these are $0,\pm\pi,\pm2\pi,\dots$ .

Knowledge used in this step

Power series and Taylor series Trigonometry Taylor's theorem