從複數到三角函數公式(II) (From complex number to trigonometric function formulas)
從複數到三角函數公式(II) (From complex number to trigonometric function formulas)
國立蘭陽女中陳敏晧教師
證明:
(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) \(\displaystyle\cos \theta+\cos 2\theta+\cdots+ \cos n\theta= \frac{{\sin \frac{{n\theta }}{2}\cos \frac{{(n + 1)\theta }}{2}}}{{\sin \frac{\theta }{2}}}\)
第二種證明方法:利用複數的概念。
我們可以使用歐拉公式 \({e^{i\theta }}= \cos \theta+ i\sin \theta\),
若將 \(\theta\) 以 \(-\theta\) 代入可得 \({e^{ – i\theta }}= \cos \theta- i\sin \theta\),
可得 \(\left\{ \begin{array}{l}\displaystyle \cos \theta= \frac{{{e^{i\theta }}+ {e^{ – i\theta }}}}{2}\\\displaystyle\sin \theta = \frac{{{e^{i\theta }}-{e^{ – i\theta }}}}{{2i}} \end{array} \right.\),變換變數得 \(\left\{ \begin{array}{l}\displaystyle\cos \frac{\theta }{2} = \frac{{{e^{\frac{{i\theta }}{2}}} + {e^{\frac{{ – i\theta }}{2}}}}}{2}\\\displaystyle\sin \frac{\theta }{2} = \frac{{{e^{\frac{{i\theta }}{2}}} – {e^{\frac{{ – i\theta }}{2}}}}}{{2i}} \end{array} \right.\)