圖形證法

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 \(\displaystyle\tan \frac{\theta }{2} = \frac{{\sin \frac{\theta }{2}}}{{\cos \frac{\theta }{2}}}=\frac{{\pm\sqrt {\frac{{1 -\cos \theta }}{2}} }}{{\pm\sqrt {\frac{{1 + \cos \theta }}{2}} }} =\pm\sqrt {\frac{{1 – \cos \theta }}{{1 + \cos \theta }}}\)，