半角公式(I)
半角公式(I) (Half-angle Formulas)
國立蘭陽女中數學科陳敏晧老師
三角函數中的半角公式:
\(\displaystyle\sin\frac{\theta}{2}=\pm \sqrt{\frac{{1-\cos\theta }}{2}}\) (\(\pm\) 號依 \(\displaystyle\frac{\theta}{2}\)在第幾象限而定)
\(\displaystyle\cos\frac{\theta}{2}=\pm \sqrt{\frac{{1+\cos\theta }}{2}}\) (\(\pm\) 號依 \(\displaystyle\frac{\theta}{2}\)在第幾象限而定)
\(\displaystyle\tan \frac{\theta }{2}=\pm\sqrt {\frac{{1-\cos \theta }}{{1+\cos \theta }}}= \frac{{\sin \theta }}{{1+ \cos \theta }}= \frac{{1- \cos \theta }}{{\sin\theta }}= \frac{{1+\sin \theta- \cos \theta }}{{1+ \sin \theta+ \cos \theta }}\)
上述半角公式的證明是根據二倍角公式:\(\cos 2\alpha= 2{\cos^2}\alpha- 1= 1- 2{\sin^2}\alpha\),
令 \(2\alpha=\theta\) 即 \(\displaystyle\alpha=\frac{\theta}{2}\),移項得 \(\displaystyle 2{\cos ^2}\frac{\theta }{2} = 1+\cos\theta ,2{\sin ^2}\frac{\theta }{2} = 1 -\cos \theta\),
再移項及開平方得 \(\displaystyle\sin \frac{\theta }{2}=\pm\sqrt{\frac{{1-\cos\theta }}{2}}\),\(\displaystyle\cos \frac{\theta }{2}=\pm\sqrt{\frac{{1+\cos\theta }}{2}}\),
將兩式相除得 \(\displaystyle\tan \frac{\theta }{2} = \frac{{\sin \frac{\theta }{2}}}{{\cos \frac{\theta }{2}}}=\frac{{\pm\sqrt {\frac{{1 -\cos \theta }}{2}} }}{{\pm\sqrt {\frac{{1 + \cos \theta }}{2}} }} =\pm\sqrt {\frac{{1 – \cos \theta }}{{1 + \cos \theta }}}\),
或根據正弦的二倍角公式 \(\sin{2\theta}\),得 \(\displaystyle\tan \frac{\theta }{2} = \frac{{\sin \frac{\theta }{2}}}{{\cos \frac{\theta }{2}}} = \frac{{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}}{{2{{\cos }^2}\frac{\theta }{2}}} = \frac{{\sin \theta }}{{1 + \cos \theta }}\),
或 \(\displaystyle\tan \frac{\theta }{2} = \frac{{\sin \frac{\theta }{2}}}{{\cos \frac{\theta }{2}}} = \frac{{2{{\sin }^2}\frac{\theta }{2}}}{{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}} = \frac{{1 – \cos \theta }}{{\sin \theta }}\),
將兩式用和比定理得 \(\begin{array}{ll}\displaystyle\tan \frac{\theta }{2}&=\displaystyle\frac{{\sin\theta }}{{1+\cos \theta }}=\frac{{1 – \cos \theta }}{{\sin\theta }}\\&=\displaystyle\frac{{\sin \theta+ (1 -\cos \theta )}}{{1+ \cos\theta+ \sin \theta }}= \frac{{1+\sin \theta- \cos \theta }}{{1 +\sin \theta+\cos \theta }}\end{array}\),
當然也可以使用分比定理得 \(\displaystyle\tan \frac{\theta }{2} = \frac{{ – 1 + \sin \theta+ \cos \theta }}{{1 – \sin \theta+ \cos \theta }}\)。
除了代數證明外,圖形證法更容易讓學生一目了然,如圖一:\(\displaystyle\tan \frac{\varphi }{2} = \frac{{\sin \varphi }}{{1 + \cos \varphi }}\)。
另一個圖形證明,如圖二:\(\displaystyle\tan\frac{\alpha }{2}=\frac{{\sin \alpha }}{{1+ \cos \alpha }}\)。
半角公式的目的除了能將三角函數降次之外,另一目的是求一些特殊角的三角函數值,如圖三:
\(\begin{array}{ll}\displaystyle\sin {11.25^\circ}&=\displaystyle\sqrt {\frac{{1 – \cos {{22.5}^\circ}}}{2}}= \frac{1}{2}\sqrt {2 – 2 \cdot \frac{1}{2}\sqrt {2 + 2\cos {{45}^\circ}} }\\&=\displaystyle \frac{1}{2}\sqrt {2 – \sqrt {2 + 2 \cdot \frac{{\sqrt 2 }}{2}} }= \frac{1}{2}\sqrt {2 – \sqrt {2 + \sqrt 2 }}\end{array}\)
\(\begin{array}{ll}\displaystyle\cos {11.25^\circ}&=\displaystyle\sqrt {\frac{{1 + \cos {{22.5}^\circ}}}{2}}= \frac{1}{2}\sqrt {2 + 2 \cdot \frac{1}{2}\sqrt {2 + 2\cos {{45}^\circ}} }\\&=\displaystyle \frac{1}{2}\sqrt {2 +\sqrt {2 + 2 \cdot \frac{{\sqrt 2 }}{2}} }= \frac{1}{2}\sqrt {2 + \sqrt {2 +\sqrt 2 } }\end{array}\)
根據上述的 \(\sin {11.25^\circ} =\frac{1}{2}\sqrt {2-\sqrt{2+\sqrt 2}}\) 值,
我們可從幾何圖形中,求出單位圓中內接正十六邊形的周長,
因為 \({360^\circ} \div 16 = {22.5^\circ}\),也就是上圖三 \(\angle BAC=22.5^\circ\),
因為 \(\overline {AD}\bot \overline {BC}\) 且 \(\overline {AB}=\overline {AC}= 1\),得 \(\angle CAD = {22.5^\circ} \div 2 = {11.25^\circ}\),
根據正弦的定義得 \(\overline {DC}=\overline {AC}\cdot \sin {11.25^\circ} = \frac{1}{2}\sqrt {2 – \sqrt {2 + \sqrt 2 } }\),
所以,單位圓中內接正十六邊形的周長為 \(16 \times 2\overline{DC}= 16\sqrt {2 – \sqrt {2 + \sqrt 2 } }\)。
連結:半角公式(II)
參考資料
- 毛爾(Eli Maor),鄭惟厚譯,《毛起來說e》,台北:天下遠見出版社,2000。
- 毛爾(Eli Maor),胡守仁譯,《毛起來說三角》,台北:天下遠見出版社,2000。
- 朱建正編撰,《數學問題百則》,台北 : 中央研究院數學研究所,1998,頁48-50。