从複数到三角函数公式(I) (From complex nu

2020-06-17 173 views

複数在数学各领域均有重大影响,本文章将讨论如何以複数的形式来证明三角函数的相关公式,由于複数具有极坐标形式,可以将角度做旋转、长度做伸缩变换,这是传统几何学在直角坐标平面难以突破的面向,因此,利用複数来证明三角函数公式往往会有意想不到的收穫,也常使学习者见识到数学之美!

本文将使用到历史法国数学家棣美弗(Abraham de Moivre, 1667-1754)于1730年发表的棣莫弗公式,即若 \(z = r(\cos \theta+ i\sin \theta)\),则 \({z^n} = {r^n}(\cos n\theta+ i\sin n\theta ),n \in Z\)。

及欧拉(Leonhard Euler, 1707-1783)在1748年所发表的欧拉公式:\({e^{i\theta }} = \cos \theta+ i\sin \theta\)。

首先论证正弦与余弦的和差角公式:

\(\sin \left( {\alpha+ \beta } \right) = \sin \alpha \cos \beta+ \cos \alpha \sin \beta\\\cos \left( {\alpha+ \beta } \right) = \cos \alpha \cos \beta- \sin \alpha \sin \beta\)

证明:\(\cos (\alpha+ \beta ) + i\sin (\alpha+ \beta )\\= {e^{i(\alpha+ \beta )}}\\= {e^{i\alpha }} \cdot {e^{i\beta }}\\= \left( {\cos \alpha+ i\sin \alpha } \right)(\cos \beta+ i\sin \beta ) \\= (\cos \alpha \cos \beta- \sin \alpha \sin \beta)+ i(\sin \alpha \cos \beta +\cos \alpha \sin \beta )\)

比较两複数的实部与虚部,即得

\(\sin \left( {\alpha+ \beta } \right) = \sin \alpha \cos \beta+ \cos \alpha \sin \beta\\\cos \left( {\alpha+ \beta } \right) = \cos \alpha \cos \beta- \sin \alpha \sin \beta\)

接着引入三倍角公式:

(1) \(\sin 3\theta= 3\sin \theta- 4{\sin ^3}\theta\) 。
(2) \(\cos 3\theta= 4{\cos ^3}\theta- 3\cos \theta\)。

证明:\(\cos 3\theta+ i\sin 3\theta\\={e^{i(3\theta )}} = {({e^{i\theta }})^3}\\= {\left( {\cos \theta+ i\sin \theta } \right)^3}\\= {(\cos \theta )^3}+3{\left( {\cos \theta}\right)^2}\cdot i\sin \theta+ 3 \cdot \cos \theta {(i\sin \theta )^2}+ {\left( {i\sin \theta}\right)^3}\\=\left( {{{\cos }^3}\theta- 3\cos \theta {{\sin }^2}\theta }\right)+i(3{\cos ^2}\theta \sin \theta- {\sin ^3}\theta )\\=\left[ {{{\cos }^3}\theta-3\cos \theta (1 – {{\cos }^2}\theta)} \right]+ i\left[{3\left( {1 – {{\sin }^2}\theta } \right)\sin \theta- {{\sin }^3}\theta } \right] \\= \left( {4{{\cos }^3}\theta- 3\cos \theta } \right)+ i(3\sin \theta- 4{\sin ^3}\theta )\)

比较两複数的实部与虚部,即得

\(\sin 3\theta= 3\sin\theta- 4{\sin ^3}\theta\\\cos 3\theta= 4{\cos ^3}\theta- 3\cos \theta\)

接着我们可以试着证明更複杂的三角函数公式:

(1) \(\displaystyle\sin \theta+\sin 2\theta+ … + \sin n\theta= \frac{{\sin \frac{{(n + 1)\theta }}{2} \cdot \sin \frac{{n\theta }}{2}}}{{\sin \frac{\theta }{2}}}\)。
(2) \(\displaystyle\cos \theta+ \cos 2\theta+ … + \cos n\theta= \frac{{\sin \frac{{n\theta }}{2}\cos \frac{{(n + 1)\theta }}{2}}}{{\sin \frac{\theta }{2}}}\)。

第一种证明方法:不用複数概念。

(1)首先考虑积化和差与和差化积的公式,即

\(2\sin \alpha \sin \beta= \cos (\alpha- \beta ) – \cos (\alpha+ \beta )\\\cos \alpha- \cos \beta= 2\sin (\frac{{\alpha+\beta }}{2})\sin \left( {\frac{{\beta-\alpha }}{2}} \right)\)

令 \(x = \sin \theta+\sin 2\theta+ … +\sin n\theta\),将等式两边同乘 \(2\sin \frac{\theta }{2}\),即得

\(2\sin \frac{\theta }{2}\cdot x\\= 2\sin \frac{\theta}{2}\cdot \sin\theta+ 2\sin \frac{\theta}{2}\cdot\sin 2\theta+ … + 2\sin \frac{\theta }{2} \cdot \sin n\theta\\= \left( {\cos \frac{\theta }{2} -\cos \frac{{3\theta }}{2}} \right)+\left( {\cos \frac{{3\theta }}{2}-\cos \frac{{5\theta }}{2}} \right)+ … + (\cos \frac{{(2n – 1)\theta }}{2}-\cos \frac{{(2n + 1)\theta }}{2})\\= \cos \frac{\theta }{2} – \cos \frac{{(2n + 1)\theta }}{2}\\= 2\sin \frac{{(n + 1)\theta }}{2} \cdot \sin \frac{{n\theta }}{2}\)

移项得 \(\displaystyle{x}=\frac{{\sin \frac{{(n + 1)\theta }}{2} \cdot \sin \frac{{n\theta }}{2}}}{{\sin \frac{\theta }{2}}}\),

即  \(\displaystyle\sin \theta+ \sin 2\theta+ … + \sin n\theta= \frac{{\sin \frac{{n\theta }}{2} \cdot \sin \frac{{(n + 1)\theta }}{2}}}{{\sin \frac{\theta }{2}}}\)。

证明:(2)首先考虑积化和差与和差化积的公式,即

\(\displaystyle 2\sin \alpha \cos \beta= \sin (\alpha+\beta)+ \sin (\alpha-\beta )\\\sin \alpha- \sin \beta= 2\sin \left( {\frac{{\alpha-\beta }}{2}}\right)\cos\left( {\frac{{\alpha+\beta }}{2}} \right)\)

令 \(y = \cos \theta+ \cos 2\theta+ … + \cos n\theta\),将等式两边同乘 \(2\sin \frac{\theta }{2}\),即得

\(2\sin \frac{\theta }{2} \cdot y\\= 2\sin \frac{\theta }{2} \cdot \cos \theta+ 2\sin \frac{\theta }{2} \cdot \cos 2\theta+\cdots+ 2\sin \frac{\theta }{2} \cdot \cos n\theta\\= (\sin \left( { – \frac{\theta }{2}} \right) + \sin \frac{{3\theta }}{2}) + (\sin ( – \frac{{3\theta }}{2}) + \sin \frac{{5\theta }}{2}) +\cdots\\~~~~~~+ \left( {\sin \left( { – \frac{{\left( {2n – 1} \right)\theta }}{2}} \right) + \sin \left( {\frac{{\left( {2n + 1} \right)\theta }}{2}} \right)} \right) \\= \sin \left( {\frac{{\left( {2n + 1} \right)\theta }}{2}} \right) – \sin \frac{\theta }{2} = 2\sin \frac{{n\theta }}{2} \cdot \cos \left( {\frac{{\left( {n + 1} \right)\theta }}{2}} \right)\)

移项得 \(y =\displaystyle\frac{{\sin \frac{{n\theta }}{2} \cdot \cos \left( {\frac{{\left( {n + 1} \right)\theta }}{2}} \right)}}{{\sin \frac{\theta }{2}}}\),

即 \(\displaystyle\cos \theta+ \cos 2\theta+\cdots+ \cos n\theta= \frac{{\sin \frac{{n\theta }}{2}\cos \frac{{(n + 1)\theta }}{2}}}{{\sin \frac{\theta }{2}}}\)。

连结:从複数到三角函数公式(II) 

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