三角函数 sinx, cosx 的泰勒展开推导及两个巧妙应用

一、推导

假设函数 fff 充分光滑，即 fff 在 x0x_0x0 点处任意阶导数存在，取一小邻域 U(x0)\small U(x_0)U(x0)，则 ∀x∈U(x0)\small \forall\,x\in U(x_0)∀x∈U(x0)，f(x)\small f(x)f(x) 都可以展开为泰勒级数，即
f(x)=∑n=0∞f(n)(x0)n!(x−x0)n=f(x0)+f′(x0)(x−x0)+f′′(x0)2!(x−x0)2+⋯f(x)=\sum_{n=0}^{\infin}\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n=f(x_0)+f'(x_0)(x-x_0)+\frac{f''(x_0)}{2!}(x-x_0)^2+\cdots f(x)=n=0∑∞n!f(n)(x0)(x−x0)n=f(x0)+f′(x0)(x−x0)+2!f′′(x0)(x−x0)2+⋯考虑三角函数 sin⁡x,cos⁡x\sin x,\cos xsinx,cosx 的任意阶导函数，
sin⁡′(x)=cos⁡(x)=sin⁡(x+π2)sin⁡′′(x)=cos⁡′(x)=sin⁡′(x+π2)=cos⁡(x+π2)=sin⁡(x+π2+π2)=sin⁡(x+π)\begin{aligned}\sin'(x)&=\cos(x)=\sin (x+\frac{\pi}{2})\\ \sin''(x)&=\cos'(x)=\sin'(x+\frac{\pi}{2})\\&=\cos (x+\frac{\pi}{2})=\sin (x+\frac{\pi}{2}+\frac{\pi}{2})\\&=\sin (x+\pi) \end{aligned}sin′(x)sin′′(x)=cos(x)=sin(x+2π)=cos′(x)=sin′(x+2π)=cos(x+2π)=sin(x+2π+2π)=sin(x+π)假设 sin⁡(n)(x)=sin⁡(x+nπ/2)\sin^{(n)}(x)=\sin(x+n\pi/2)sin(n)(x)=sin(x+nπ/2)，下面我们用 数学归纳法 来验证一下.

首先，当 n=0n=0n=0 时，sin⁡(0)(x)=sin⁡(x)\sin^{(0)}(x)=\sin(x)sin(0)(x)=sin(x)，结论显然成立.

其次，假设当 n=k(k≥0)n=k(k\geq 0)n=k(k≥0) 时结论成立，即 sin⁡(k)(x)=sin⁡(x+kπ/2)\displaystyle\sin^{(k)}(x)=\sin(x+k\pi/2)sin(k)(x)=sin(x+kπ/2).

考虑 n=k+1n=k+1n=k+1 的情形，此时
sin⁡(k+1)(x)=d(sin⁡(k)(x))dx=d(sin⁡(x+kπ/2))dx=cos⁡(x+kπ/2)=sin⁡(x+kπ/2+π/2)=sin⁡(x+(k+1)π/2)\begin{aligned} \sin^{(k+1)}(x)&=\frac{d(\sin^{(k)}(x))}{dx}\\&=\frac{d(\sin(x+k\pi/2))}{dx}\\&=\cos(x+k\pi/2)\\&=\sin(x+k\pi/2+\pi/2)\\&=\sin(x+(k+1)\pi/2) \end{aligned} sin(k+1)(x)=dxd(sin(k)(x))=dxd(sin(x+kπ/2))=cos(x+kπ/2)=sin(x+kπ/2+π/2)=sin(x+(k+1)π/2) 所以，当 n=k+1n=k+1n=k+1 时，结论照样成立.

由归纳法原理，∀n∈N,sin⁡(n)(x)=sin⁡(x+nπ/2)\forall\,n\in N,\,\sin^{(n)}(x)=\sin(x+n\pi/2)∀n∈N,sin(n)(x)=sin(x+nπ/2).

则 cos⁡(n)(x)=sin⁡(n+1)(x)=sin⁡(x+nπ/2+π/2)=cos⁡(x+nπ/2)\cos^{(n)}(x)=\sin^{(n+1)}(x)=\sin(x+n\pi/2+\pi/2)=\cos(x+n\pi/2)cos(n)(x)=sin(n+1)(x)=sin(x+nπ/2+π/2)=cos(x+nπ/2).

特别地，取 x=0x=0x=0，

sin⁡(2k)(0)=sin⁡(kπ)=0sin⁡(2k+1)(0)=sin⁡(kπ+π/2)=(−1)kcos⁡(2k)(0)=cos⁡(kπ)=(−1)kcos⁡(2k+1)(0)=cos⁡(kπ+π/2)=0(k=0,1,2,⋯)\begin{aligned} &\sin^{(2k)}(0)=\sin(k\pi)=0 \\ &\sin^{(2k+1)}(0)=\sin(k\pi+\pi/2)=(-1)^{k} \\ \\ &\cos^{(2k)}(0)=\cos(k\pi)=(-1)^{k} \\ &\cos^{(2k+1)}(0)=\cos(k\pi+\pi/2)=0 \\ &(k=0,1,2,\cdots) \end{aligned} sin(2k)(0)=sin(kπ)=0sin(2k+1)(0)=sin(kπ+π/2)=(−1)kcos(2k)(0)=cos(kπ)=(−1)kcos(2k+1)(0)=cos(kπ+π/2)=0(k=0,1,2,⋯) 取 x0=0x_0=0x0=0 的某一小邻域，将 sin⁡x,cos⁡x\sin x,\cos xsinx,cosx 泰勒展开
sin⁡x=∑n=0∞sin⁡(n)(0)n!xn=∑k=0∞sin⁡(2k)(0)(2k)!x2k+∑k=0∞sin⁡(2k+1)(0)(2k+1)!x2k+1=∑k=0∞0(2k)!x2k+∑k=0∞(−1)k(2k+1)!x2k+1=∑k=0∞(−1)k(2k+1)!x2k+1=x−13!x3+15!x5−17!x7+⋯cos⁡x=∑n=0∞cos⁡(n)(0)n!xn=∑k=0∞cos⁡(2k)(0)(2k)!x2k+∑k=0∞cos⁡(2k+1)(0)(2k+1)!x2k+1=∑k=0∞(−1)k(2k)!x2k+∑k=0∞0(2k+1)!x2k+1=∑k=0∞(−1)k(2k)!x2k=1−12!x2+14!x4−16!x6+⋯\begin{aligned} \sin x&=\sum_{n=0}^{\infin}\frac{\sin^{(n)}(0)}{n!}x^n\\ &=\sum_{k=0}^{\infin}\frac{\sin^{(2k)}(0)}{(2k)!}x^{2k}+\sum_{k=0}^{\infin}\frac{\sin^{(2k+1)}(0)}{(2k+1)!}x^{2k+1}\\ &=\sum_{k=0}^{\infin}\frac{0}{(2k)!}x^{2k}+\sum_{k=0}^{\infin}\frac{(-1)^k}{(2k+1)!}x^{2k+1}\\ &=\sum_{k=0}^{\infin}\frac{(-1)^k}{(2k+1)!}x^{2k+1}\\ &=x-\frac{1}{3!}x^3+\frac{1}{5!}x^5-\frac{1}{7!}x^7+\cdots\\ \\ \cos x&=\sum_{n=0}^{\infin}\frac{\cos^{(n)}(0)}{n!}x^n\\ &=\sum_{k=0}^{\infin}\frac{\cos^{(2k)}(0)}{(2k)!}x^{2k}+\sum_{k=0}^{\infin}\frac{\cos^{(2k+1)}(0)}{(2k+1)!}x^{2k+1}\\ &=\sum_{k=0}^{\infin}\frac{(-1)^{k}}{(2k)!}x^{2k}+\sum_{k=0}^{\infin}\frac{0}{(2k+1)!}x^{2k+1}\\ &=\sum_{k=0}^{\infin}\frac{(-1)^{k}}{(2k)!}x^{2k}\\ &=1-\frac{1}{2!}x^2+\frac{1}{4!}x^4-\frac{1}{6!}x^6+\cdots \end{aligned} sinxcosx=n=0∑∞n!sin(n)(0)xn=k=0∑∞(2k)!sin(2k)(0)x2k+k=0∑∞(2k+1)!sin(2k+1)(0)x2k+1=k=0∑∞(2k)!0x2k+k=0∑∞(2k+1)!(−1)kx2k+1=k=0∑∞(2k+1)!(−1)kx2k+1=x−3!1x3+5!1x5−7!1x7+⋯=n=0∑∞n!cos(n)(0)xn=k=0∑∞(2k)!cos(2k)(0)x2k+k=0∑∞(2k+1)!cos(2k+1)(0)x2k+1=k=0∑∞(2k)!(−1)kx2k+k=0∑∞(2k+1)!0x2k+1=k=0∑∞(2k)!(−1)kx2k=1−2!1x2+4!1x4−6!1x6+⋯则
sin⁡x=∑k=0∞(−1)k(2k+1)!x2k+1=x−13!x3+15!x5−17!x7+⋯cos⁡x=∑k=0∞(−1)k(2k)!x2k=1−12!x2+14!x4−16!x6+⋯\begin{aligned} &\sin x=\sum_{k=0}^{\infin}\frac{(-1)^k}{(2k+1)!}x^{2k+1} =x-\frac{1}{3!}x^3+\frac{1}{5!}x^5-\frac{1}{7!}x^7+\cdots \\ &\cos x=\sum_{k=0}^{\infin}\frac{(-1)^{k}}{(2k)!}x^{2k} =1-\frac{1}{2!}x^2+\frac{1}{4!}x^4-\frac{1}{6!}x^6+\cdots \end{aligned} sinx=k=0∑∞(2k+1)!(−1)kx2k+1=x−3!1x3+5!1x5−7!1x7+⋯cosx=k=0∑∞(2k)!(−1)kx2k=1−2!1x2+4!1x4−6!1x6+⋯下面打算用这两个式子证些好玩儿的东西.

二、两个妙用

1. 欧拉公式

eiθ=cos⁡θ+isin⁡θe^{i\theta}=\cos\theta+i\sin\theta eiθ=cosθ+isinθ 首先将 exe^xex 在 x0=0x_0=0x0=0 处进行泰勒展开，已知 (ex)′=ex(e^x)'=e^x(ex)′=ex，则 (ex)(n)∣x=0=ex∣x=0=1(e^x)^{(n)}|_{x=0}=e^x|_{x=0}=1(ex)(n)∣x=0=ex∣x=0=1，所以
ex=∑n=0∞(ex)(n)∣x=0n!xn=∑n=0∞1n!xn=1+x+12!x2+13!x3+⋯e^x=\sum_{n=0}^{\infin}\frac{(e^x)^{(n)}|_{x=0}}{n!}x^n=\sum_{n=0}^{\infin}\frac{1}{n!}x^n=1+x+\frac{1}{2!}x^2+\frac{1}{3!}x^3+\cdots ex=n=0∑∞n!(ex)(n)∣x=0xn=n=0∑∞n!1xn=1+x+2!1x2+3!1x3+⋯将 xxx 用 iθi\thetaiθ 代替，得
eiθ=1+iθ−12!θ2−i3!θ3+14!θ4+i5!θ5−16!θ6−i7!θ7+⋯=(1−12!θ2+14!θ4−16θ6+⋯)+i(θ−13!θ3+15!θ5−17!θ7+⋯)\begin{aligned} e^{i\theta}&=1+i\theta-\frac{1}{2!}\theta^2-\frac{i}{3!}\theta^3+\frac{1}{4!}\theta^4+\frac{i}{5!}\theta^5-\frac{1}{6!}\theta^6-\frac{i}{7!}\theta^7+\cdots\\ &=(1-\frac{1}{2!}\theta^2+\frac{1}{4!}\theta^4-\frac{1}{6}\theta^6+\cdots)+i(\theta-\frac{1}{3!}\theta^3+\frac{1}{5!}\theta^5-\frac{1}{7!}\theta^7+\cdots) \end{aligned} eiθ=1+iθ−2!1θ2−3!iθ3+4!1θ4+5!iθ5−6!1θ6−7!iθ7+⋯=(1−2!1θ2+4!1θ4−61θ6+⋯)+i(θ−3!1θ3+5!1θ5−7!1θ7+⋯)将1−12!θ2+14!θ4−16θ6+⋯=cos⁡θθ−13!θ3+15!θ5−17!θ7+⋯=sin⁡θ\begin{aligned} 1-\frac{1}{2!}\theta^2+\frac{1}{4!}\theta^4-\frac{1}{6}\theta^6+\cdots=\cos\theta\\ \theta-\frac{1}{3!}\theta^3+\frac{1}{5!}\theta^5-\frac{1}{7!}\theta^7+\cdots=\sin\theta \end{aligned} 1−2!1θ2+4!1θ4−61θ6+⋯=cosθθ−3!1θ3+5!1θ5−7!1θ7+⋯=sinθ代入，得
eiθ=cos⁡θ+isin⁡θe^{i\theta}=\cos\theta+i\sin\theta eiθ=cosθ+isinθ特别地，考虑 θ=π\theta=\piθ=π 时，
eiπ+1=0e^{i\pi}+1=0 eiπ+1=0这是一个奇妙的公式，同时包含了数学中的 5 个重要的常数：e,i,π,1,0e,i,\pi,1,0e,i,π,1,0.

另外， eiθe^{i\theta}eiθ 的模长为1，常用于积分换元. 证明一下:

eiθ‾=eiθ‾=e−iθ\overline {e^{i\theta}}=e^{\overline{i\theta}}=e^{-i\theta}eiθ=eiθ=e−iθ∴∣eiθ∣2=eiθ⋅eiθ‾=eiθ−iθ=e0=1\therefore\,\, |e^{i\theta}|^2=e^{i\theta}\cdot\overline {e^{i\theta}}=e^{i\theta-i\theta}=e^0=1 ∴∣eiθ∣2=eiθ⋅eiθ=eiθ−iθ=e0=1
eiθ=cos⁡θ+isin⁡θe^{i\theta}=\cos\theta+i\sin\thetaeiθ=cosθ+isinθ∴∣eiθ∣2=cos⁡θcos⁡θ+sin⁡θsin⁡θ=cos⁡(θ−θ)=cos⁡0=1\therefore\,\, |e^{i\theta}|^2=\cos\theta\cos\theta+\sin\theta\sin\theta=\cos(\theta-\theta)=\cos 0 =1 ∴∣eiθ∣2=cosθcosθ+sinθsinθ=cos(θ−θ)=cos0=1

2. 自然数倒数平方和

∑n=1∞1n2=1+122+132+142+⋯=π26\sum_{n=1}^{\infin}\frac{1}{n^2}=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots=\frac{\pi^2}{6} n=1∑∞n21=1+221+321+421+⋯=6π2 虽然不叫欧拉公式，但这种解法同样是欧拉给出来的，下面让我们欣赏下欧拉的伟大思路.

考虑函数
f(x)=sin⁡xx(x≠0),lim⁡x→0f(x)=1f(x)=\frac{\sin x}{x}(x\neq 0),\,\,\lim_{x\to 0}f(x)=1 f(x)=xsinx(x=0),x→0limf(x)=1补充定义 f(0)=1f(0)=1f(0)=1.

因为 f(x)=sin⁡x/xf(x)=\sin x/xf(x)=sinx/x，所以其零点为 kπ,k=±1,±2,±3,⋯k\pi,k=\pm1,\pm2,\pm3,\cdotskπ,k=±1,±2,±3,⋯

将 f(x)f(x)f(x) 看作一个多项式，可以通过零点表示，即
f(x)=c1(x−π)(x+π)(x−2π)(x+2π)(x−3π)(x+3π)⋯=c1(x2−π2)(x2−4π2)(x2−9π2)⋯\begin{aligned} f(x)&=c_1(x-\pi)(x+\pi)(x-2\pi)(x+2\pi)(x-3\pi)(x+3\pi)\cdots\\ &=c_1(x^2-\pi^2)(x^2-4\pi^2)(x^2-9\pi^2)\cdots \end{aligned} f(x)=c1(x−π)(x+π)(x−2π)(x+2π)(x−3π)(x+3π)⋯=c1(x2−π2)(x2−4π2)(x2−9π2)⋯将
sin⁡x=∑k=0∞(−1)k(2k+1)!x2k+1=x−13!x3+15!x5−17!x7+⋯\sin x=\sum_{k=0}^{\infin}\frac{(-1)^k}{(2k+1)!}x^{2k+1} =x-\frac{1}{3!}x^3+\frac{1}{5!}x^5-\frac{1}{7!}x^7+\cdots sinx=k=0∑∞(2k+1)!(−1)kx2k+1=x−3!1x3+5!1x5−7!1x7+⋯代入 f(x)=sin⁡x/xf(x)=\sin x/xf(x)=sinx/x，得
f(x)=1−13!x2+15!x4−17!x6+⋯f(x)=1-\frac{1}{3!}x^2+\frac{1}{5!}x^4-\frac{1}{7!}x^6+\cdots f(x)=1−3!1x2+5!1x4−7!1x6+⋯将这两个式子放在一起
1−13!x2+15!x4−17!x6+⋯=c1(x2−π2)(x2−4π2)(x2−9π2)⋯1-\frac{1}{3!}x^2+\frac{1}{5!}x^4-\frac{1}{7!}x^6+\cdots=c_1(x^2-\pi^2)(x^2-4\pi^2)(x^2-9\pi^2)\cdots 1−3!1x2+5!1x4−7!1x6+⋯=c1(x2−π2)(x2−4π2)(x2−9π2)⋯由两端常数项相等，得
1=c1(−π2)(−4π2)(−9π2)⋯c1=1(−π2)(−4π2)(−9π2)⋯1=c_1(-\pi^2)(-4\pi^2)(-9\pi^2)\cdots\\ c_1=\frac{1}{(-\pi^2)(-4\pi^2)(-9\pi^2)\cdots} 1=c1(−π2)(−4π2)(−9π2)⋯c1=(−π2)(−4π2)(−9π2)⋯1代入原式，得
1−13!x2+15!x4−17!x6+⋯=(1−x2π2)(1−x24π2)(1−x29π2)⋯1-\frac{1}{3!}x^2+\frac{1}{5!}x^4-\frac{1}{7!}x^6+\cdots=(1-\frac{x^2}{\pi^2})(1-\frac{x^2}{4\pi^2})(1-\frac{x^2}{9\pi^2})\cdots 1−3!1x2+5!1x4−7!1x6+⋯=(1−π2x2)(1−4π2x2)(1−9π2x2)⋯
考虑两端 x2x^2x2 的系数，有人可能会问右端 x2x^2x2 的系数是怎么求出来的？没错，我也想问.

我的理解是这样的：利用组合的思想，右端的 x2x^2x2 项只能这样得出：第 kkk 个括号里的 −x2k2π2\displaystyle-\frac{x^2}{k^2\pi^2}−k2π2x2 与其他括号中的 111 相乘，然后将这无穷多项相加.

所以右端 x2x^2x2 的系数为
−1π2−14π2−19π2−⋯=−1π2∑n=1∞1n2-\frac{1}{\pi^2}-\frac{1}{4\pi^2}-\frac{1}{9\pi^2}-\cdots=-\frac{1}{\pi^2}\sum_{n=1}^{\infin}\frac{1}{n^2} −π21−4π21−9π21−⋯=−π21n=1∑∞n21而左端 x2x^2x2 的系数为 −13!\displaystyle -\frac{1}{3!}−3!1，所以
−13!=−1π2∑n=1∞1n2π26=∑n=1∞1n2=1+122+132+142+⋯\begin{aligned} -\frac{1}{3!}&=-\frac{1}{\pi^2}\sum_{n=1}^{\infin}\frac{1}{n^2}\\ \frac{\pi^2}{6}&=\sum_{n=1}^{\infin}\frac{1}{n^2}=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots \end{aligned} −3!16π2=−π21n=1∑∞n21=n=1∑∞n21=1+221+321+421+⋯Okay, 证明完毕！

如有纰漏，还请指正！