# 常用三角函数的无穷级数乘积公式推导详细过程与图形展示

## 多项式逼近理论得到常用三角函数的无穷级数乘积公式

### 问题: 求方程 sin ⁡ ( x ) = 0 \sin(x)=0 sin(x)=0 的解。

1、首先, sin ⁡ ( x ) = 0 \sin(x)=0 sin(x)=0 有解 { k π ， k = 0 , ± 1 , ± 2 , ⋯   . } \{kπ，k=0,\pm 1,\pm 2, \cdots .\} { kπk=0,±1,±2,.}

2、假设 sin ⁡ ( x ) \sin(x) sin(x) 是多项式函数，由多项式有根的代数基本理论 (the fundamental theorem of algebra) 即 多项式逼近理论 (polynomial approximation theorem)

sin ⁡ ( x ) = c ∗ ∏ k = 1 ∞ ( k π − x ) ( k π + x ) x sin ⁡ ( x ) x = c ∗ ∏ k = 1 ∞ ( k π − x ) ( k π + x ) \displaystyle \sin(x) = c*\prod_{k=1}^{\infty} (kπ-x)(kπ+x)x \\[1em]\frac{\sin(x)}{x} = c*\prod_{k=1}^{\infty} (kπ-x)(kπ+x) sin(x)=ck=1(kπx)(kπ+x)xxsin(x)=ck=1(kπx)(kπ+x)

x → 0 x \to 0 x0， 求极限得到
c = ∏ k = 1 ∞ 1 k 2 π 2 \displaystyle c=\prod_{k=1}^{\infty} \frac {1}{k^2π^2} c=k=1k2π21

f s i n ( x ) = sin ⁡ ( x ) = x ∏ k = 1 ∞ ( 1 − x k π ) ( 1 + x k π ) = x ∏ k = 1 ∞ [ 1 − x 2 ( k π ) 2 ] \displaystyle \begin{aligned} \red {fsin(x)} =\sin(x) &= x {\prod_{k=1} ^{\infty} ({1- \frac{x}{kπ})} (1+ \frac{x}{kπ})} \\&= x \prod_{k=1} ^{\infty} \left[1- \frac{x^2}{(kπ)^2}\right ] \end{aligned} fsin(x)=sin(x)=xk=1(1kπx)(1+kπx)=xk=1[1(kπ)2x2]

x = π 2 x = \frac{π}{2} x=2π, 有

π 2 = ∏ k = 1 ∞ 2 k 2 k − 1 ⋅ 2 k 2 k + 1 = ∏ k = 1 ∞ 1 1 − 1 4 k 2 = ∏ k = 1 ∞ [ 1 + 1 4 k 2 − 1 ] = 2 1 ⋅ 2 3 ⋅ 4 3 ⋅ 4 5 ⋅ 6 5 ⋅ 6 7 ⋯ = 4 3 ⋅ 16 15 ⋅ 36 35 ⋅ 64 63 ⋯ \displaystyle \begin{aligned} \frac{π}{2} &= \prod_{k=1} ^{\infty} \frac{2k}{2k-1} \centerdot \frac{2k}{2k+1} \\&= \prod_{k=1} ^{\infty} \frac{1}{1- \frac{1}{4k^2}} \\&= \prod_{k=1} ^{\infty} \left[1+\frac{1}{4k^2-1}\right] \\&=\frac{2}{1} \centerdot \frac{2}{3} \centerdot \frac{4}{3} \centerdot \frac{4}{5} \centerdot \frac{6}{5} \centerdot \frac{6}{7} \cdots \\&=\frac{4}{3} \centerdot \frac{16}{15} \centerdot \frac{36}{35} \centerdot \frac{64}{63} \cdots \end{aligned} 2π=k=12k12k2k+12k=k=114k211=k=1[1+4k211]=123234545676=34151635366364

f c 1 ( x ) = cos ⁡ ( x ) = ( π 2 − x ) ∏ k = 1 ∞ ( 1 − π / 2 − x k π ) ( 1 + π / 2 − x k π ) = ( π 2 − x ) ∏ k = 1 ∞ ( 1 − 1 2 k + x k π ) ( 1 + 1 2 k − x k π ) \displaystyle \begin{aligned} \red{fc_1(x)}=\cos(x)&=(\frac{\pi}{2}-x) \prod_{k=1} ^{\infty} {\left(1 – \frac{\pi/2-x}{kπ}\right)} {\left(1 + \frac{\pi/2-x}{kπ}\right)}\\&=(\frac{\pi}{2}-x) \prod_{k=1} ^{\infty} {\left(1-\frac{1}{2k}+\frac{x}{k \pi}\right)}{\left(1+\frac{1}{2k}-\frac{x}{k \pi}\right)} \end{aligned} fc1(x)=cos(x)=(2πx)k=1(1kππ/2x)(1+kππ/2x)=(2πx)k=1(12k1+kπx)(1+2k1kπx)

f c o s ( x ) = cos ⁡ ( x ) = ( 1 − 2 x π ) ∏ k = 1 ∞ ( 1 + 2 x ( 2 k − 1 ) π ) ( 1 − 2 x ( 2 k + 1 ) π ) = ( 1 − 2 x π ) ∏ k = 1 ∞ ( 1 + x / π k − 1 / 2 ) ( 1 − x / π k + 1 / 2 ) \displaystyle \begin{aligned} \red{fcos(x)}=\cos(x)&=\left(1-\frac{2x}{\pi}\right) \prod_{k=1} ^{\infty} {\left(1 + \frac{2x}{(2k-1)π}\right)} {\left(1 – \frac{2x}{(2k+1)π}\right)}\\&=\left(1-\frac{2x}{\pi}\right) \prod_{k=1} ^{\infty} {\left(1+\frac{x/\pi}{k-1/2}\right)}{\left(1-\frac{x/\pi}{k+1/2}\right)} \end{aligned} fcos(x)=cos(x)=(1π2x)k=1(1+(2k1)π2x)(1(2k+1)π2x)=(1π2x)k=1(1+k1/2x/π)(1k+1/2x/π)

f t a n ( x ) = tan ⁡ ( x ) = 4 x π ⋅ ∏ k = 1 ∞ 1 − x k π 1 − 1 4 k ⋅ 1 + x k π 1 + 1 4 k \displaystyle \begin{aligned} \red{ftan(x)}= \tan(x)&=\frac{4x}{\pi} \centerdot \prod_{k=1} ^{\infty} \frac{1 – \frac{x}{kπ}}{1-\frac{1}{4k}} \centerdot \frac{1 + \frac{x}{kπ}}{1+\frac{1}{4k}}\end{aligned} ftan(x)=tan(x)=π4xk=114k11kπx1+4k11+kπx

原文作者：学而思，思而在！讷于言，敏于行！
原文地址: https://blog.csdn.net/liuxiang3/article/details/120015470
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