---
tag: 6M016JBDF0
kind: proof
name: Proof of the power series of sine
msc: [33B10, 30B10]
proves: 02KTRHQMHN
uses: [0WB57D9488, 0QNY6ZBXPN, 0W3K3QPEWT, 00R4MWW532, 0V2DV29EGT, 084NGSTWM7, 043SJ5RF59, 0WFCGVMRRC]
sources:
  - url: https://en.wikipedia.org/wiki/Sine_and_cosine
    locator: Series and polynomials
---

## Proof

1. Fix a real $x$. Let $u_n = \frac{(-1)^n x^{2n+1}}{(2n+1)!}$ for $n \ge 0$, where $k!$ is the [[factorial|0WFCGVMRRC]].
2. If $x = 0$, every $u_n$ is $0$, so $\sum \abs{u_n}$ converges.
3. If $x \ne 0$, then $u_n \ne 0$ and $\frac{\abs{u_{n+1}}}{\abs{u_n}} = \frac{x^2}{(2n+2)(2n+3)} \to 0 < 1$.
4. By the [[ratio test|0QNY6ZBXPN]], $\sum u_n$ converges [[absolutely|0W3K3QPEWT]] in this case too.
5. By the [[definition of sine|0WB57D9488]], $\sin x = \sum_{n=0}^{\infty} u_n = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$.
6. It remains to identify this series as the Maclaurin series. By the [[derivative of sine|00R4MWW532]] and the [[derivative of cosine|0V2DV29EGT]], $\sin' = \cos$ and $\cos' = -\sin$ on $\mathbb{R}$.
7. By induction on $k$, the [[higher derivatives|084NGSTWM7]] exist for all orders: $\sin^{(4k)} = \sin$, $\sin^{(4k+1)} = \cos$, $\sin^{(4k+2)} = -\sin$ and $\sin^{(4k+3)} = -\cos$. So $\sin$ is smooth.
8. At $0$, every term of the sine series vanishes, and only the first term of the cosine series is nonzero. So $\sin 0 = 0$ and $\cos 0 = 1$.
9. By steps 7 and 8, $\sin^{(m)}(0) = 0$ for even $m$, $\sin^{(m)}(0) = 1$ for $m = 4k + 1$, and $\sin^{(m)}(0) = -1$ for $m = 4k + 3$.
10. So for $m = 2n + 1$ we get $\sin^{(m)}(0) = (-1)^n$, since $n$ is even exactly when $m = 4k + 1$.
11. The [[Maclaurin series|043SJ5RF59]] $\sum_m \frac{\sin^{(m)}(0)}{m!} x^m$ thus has zero terms for even $m$, and the term $\frac{(-1)^n x^{2n+1}}{(2n+1)!}$ for $m = 2n + 1$.
12. Removing the zero terms, the Maclaurin series of $\sin$ is $\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}$, the series of the statement.

## Idea

The terms shrink like $\frac{1}{(2n+1)!}$, which beats any power of $x$, so the ratio test gives absolute convergence. With sine defined by this series, the equality is the definition. The derivatives of $\sin$ at $0$ cycle through $0, 1, 0, -1$, and these values give back the same coefficients.