从複数到三角函数公式(II) (From complex n

2020-06-17 401 views

连结:从複数到三角函数公式(I) 

证明:

(1) \(\displaystyle\sin \theta+ \sin 2\theta+\cdots+ \sin n\theta= \frac{{\sin \frac{{(n + 1)\theta }}{2} \cdot \sin \frac{{n\theta }}{2}}}{{\sin \frac{\theta }{2}}}\)

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

第二种证明方法:利用複数的概念。

我们可以使用欧拉公式 \({e^{i\theta }}= \cos \theta+ i\sin \theta\),

若将 \(\theta\) 以 \(-\theta\) 代入可得 \({e^{ – i\theta }}= \cos \theta- i\sin \theta\),

可得 \(\left\{ \begin{array}{l}\displaystyle \cos \theta= \frac{{{e^{i\theta }}+ {e^{ – i\theta }}}}{2}\\\displaystyle\sin \theta = \frac{{{e^{i\theta }}-{e^{ – i\theta }}}}{{2i}} \end{array} \right.\),变换变数得 \(\left\{ \begin{array}{l}\displaystyle\cos \frac{\theta }{2} = \frac{{{e^{\frac{{i\theta }}{2}}} + {e^{\frac{{ – i\theta }}{2}}}}}{2}\\\displaystyle\sin \frac{\theta }{2} = \frac{{{e^{\frac{{i\theta }}{2}}} – {e^{\frac{{ – i\theta }}{2}}}}}{{2i}} \end{array} \right.\)

,将上式令为

\(\begin{array}{ll}y+ix&= (\cos \theta+\cos 2\theta+\cdots+ \cos n\theta )+ i(\sin \theta+\sin 2\theta+\cdots+ \sin n\theta )\\&= (\cos \theta+ i\sin \theta )+ (\cos 2\theta+ i\sin 2\theta )+\cdots+ (\cos n\theta+ i\sin n\theta )\\&= (\cos \theta+i\sin \theta )+{(\cos \theta+i\sin \theta )^2}+\cdots+ {(\cos \theta+ i\sin \theta )^n}\\&= {e^{i\theta }}+{e^{i2\theta }} +\cdots+ {e^{in\theta }}\end{array}\)
\(\begin{array}{ll}~~~~~~~~&=\frac{{{e^{i\theta }} \cdot (1 – {e^{in\theta }})}}{{1 – {e^{i\theta }}}}\end{array}\),分母 \(1 – {e^{i\theta }} = {e^{\frac{{i\theta }}{2}}} \cdot ({e^{\frac{{ – i\theta }}{2}}} – {e^{\frac{{i\theta }}{2}}}) = {e^{\frac{{i\theta }}{2}}}( – 2i\sin \frac{\theta }{2})\)
\(\begin{array}{ll}~~~~~~~~&=\displaystyle\frac{{{e^{i\theta }} \cdot (1 – {e^{in\theta }})}}{{{e^{\frac{{i\theta }}{2}}}( – 2i\sin \frac{\theta }{2})}}\\&=\displaystyle \frac{{{e^{\frac{{i\theta }}{2}}} – {e^{\frac{{i(2n + 1)\theta }}{2}}}}}{{ – 2i\sin \frac{\theta }{2}}}\\&=\displaystyle\frac{{\cos \frac{\theta }{2} + i\sin \frac{\theta }{2} – \cos \frac{{(2n + 1)\theta }}{2} – i\sin \frac{{\left( {2n + 1} \right)\theta }}{2}}}{{ – 2i\sin \frac{\theta }{2}}}\\&=\displaystyle\frac{{(\sin \frac{{(2n + 1)\theta }}{2} – \sin \frac{\theta }{2}) + i(\cos \frac{\theta }{2} – \cos \frac{{(2n + 1)\theta }}{2})}}{{2\sin \frac{\theta }{2}}}\end{array}\)

因此,\(\displaystyle{y}= \cos \theta+ \cos 2\theta+\cdots+ \cos n\theta= \frac{{\sin \frac{{(2n + 1)\theta }}{2}-\sin \frac{\theta }{2}}}{{2\sin \frac{\theta }{2}}}=\frac{{\sin \frac{{n\theta }}{2}\cos \frac{{(n + 1)\theta }}{2}}}{{\sin \frac{\theta }{2}}}\\\displaystyle{x}= \sin \theta+\sin 2\theta+\cdots+\sin n\theta= \frac{{\cos \frac{\theta }{2}- \cos \frac{{(2n + 1)\theta }}{2}}}{{2\sin \frac{\theta }{2}}}=\frac{{\sin \frac{{n\theta }}{2} \cdot \sin \frac{{(n + 1)\theta }}{2}}}{{\sin \frac{\theta }{2}}}\)

第三种证明方法:图形証法。

从複数到三角函数公式(II) (From complex n

如上图所示,我们令 \(\overline{BB_1}=1\),且 \(\overline{BB_1}\) 与 \(x\) 轴之夹角为 \(\theta\),

再过点 \(B_1\) 做另一条线段 \(\overline{B_1B_2}=1\),并使其与第一条线段 \(\overline{BB_1}\) 的夹角为 \(\theta\),

所以,\(\overline{B_1B_2}\) 与 \(x\) 轴之夹角为 \(2\theta\),以此类推做 \(n\) 次,得到点 \(B_n\),

利用对同弧时圆周角等于其所对圆心角的性质,

则 \(B,B_1,B_2,\cdots,B_n\) 都在以 \(A\) 为圆心的圆周上,

令 \(B\left( {0,0} \right),{B_1}\left( {\cos \theta ,\sin \theta } \right),{B_2}(\cos \theta+\cos 2\theta ,\sin \theta+\sin 2\theta )\),以此类推,

令 \({B_n}(\cos \theta+ \cos 2\theta+\cdots+ \cos n\theta ,\sin \theta+\sin 2\theta+\cdots+\sin n\theta )\),

在 \(\Delta ABB_1\) 中,因为 \(\overline{AB}=\overline{AB_1}\),所以 \(\angle AB{B_1} = \angle A{B_1}B = {90^\circ} – \frac{\theta }{2}\),

根据正弦定理,即 \(\displaystyle\frac{{\overline {AB} }}{{\sin ({{90}^\circ} – \frac{\theta }{2})}} = \frac{{\overline {B{B_1}} }}{{\sin \theta }}\),

利用正弦二倍角公式,得到 \(\displaystyle\overline {AB}= \frac{{1 \cdot \cos \frac{\theta }{2}}}{{\sin \theta }} = \frac{{\cos \frac{\theta }{2}}}{{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}} = \frac{1}{{2\sin \frac{\theta }{2}}}\),

接着考虑 \(\Delta AB{B_n}\),因为顶角 \(\angle BA{B_n}= n\theta\),则底角 \(\angle AB{B_n} = \angle A{B_n}B = {90^\circ} – \frac{{n\theta }}{2}\),

再根据正弦定理及正弦二倍角公式 \(\displaystyle\frac{{\overline {AB} }}{{\sin ({{90}^\circ} – \frac{{n\theta }}{2})}} = \frac{{\overline {B{B_n}} }}{{\sin n\theta }}\),

化简得 \(\displaystyle\overline {BB_n}= \overline {AB}\cdot \frac{{\sin n\theta }}{{\cos \frac{{n\theta }}{2}}}= \frac{1}{{2\sin \frac{\theta }{2}}} \cdot \frac{{2\sin \frac{{n\theta }}{2} \cdot \cos \frac{{n\theta }}{2}}}{{\cos \frac{{n\theta }}{2}}}= \frac{{\sin \frac{{n\theta }}{2}}}{{\sin \frac{\theta }{2}}}\),

利用圆周角性质 \(\angle {B_1}B{B_n} = \frac{1}{2}\angle {B_1}A{B_n} = \frac{{(n – 1)\theta }}{2}\),即图上 \(\alpha= \frac{{(n – 1)\theta }}{2}\),

算得 \(\alpha+ \theta= \frac{{\left( {n – 1} \right)\theta }}{2}+ \theta= \frac{{(n + 1)\theta }}{2}\),

最后根据正弦与余弦定义得

\(\displaystyle\cos \theta+ \cos 2\theta+\cdots+\cos n\theta= \overline {B{B_n}}\cdot \cos \frac{{(n + 1)\theta }}{2}= \frac{{\sin \frac{{n\theta }}{2} \cdot \cos \frac{{(n + 1)\theta }}{2}}}{{\sin \frac{\theta }{2}}}\)

\(\displaystyle\sin \theta+ \sin 2\theta+\cdots+ \sin n\theta= \overline {B{B_n}}\cdot \sin \frac{{(n + 1)\theta }}{2}= \frac{{\sin \frac{{n\theta }}{2} \cdot \sin \frac{{(n + 1)\theta }}{2}}}{{\sin \frac{\theta }{2}}}\)

欧拉公式 \({e^{i\pi }} + 1 = 0\) 常常被誉为数学界中最美的公式。

最后,笔者引《博士热爱的算式》第163页做为结语:

欧拉从这些看似毫无关係的数字中,发现彼此之间自然的关联:「\(e\) 的 \(\pi\) 和 \(i\) 之积的次方再加上 \(1\),就变成了 \(0\)。」我重新看着博士的纸条。永无止境循环下去的数字,和让人难以捉摸的虚数画出简洁的轨迹,在某一点落地。虽然没有圆的出现,但来自宇宙的 \(\pi\) 飘然来到 \(e\) 的身旁,和害羞的 \(i\) 握着手。他们的身体紧紧靠在一起,屏住呼吸;但有人加了 \(1\) 以后,世界就毫无预警发生了巨大的变化,一切都归于 \(0\)。

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