半角公式(I)

发布于:2020-06-20 分类:E猫生活   

三角函数中的半角公式:

\(\displaystyle\sin\frac{\theta}{2}=\pm \sqrt{\frac{{1-\cos\theta }}{2}}\) (\(\pm\) 号依 \(\displaystyle\frac{\theta}{2}\)在第几象限而定)

\(\displaystyle\cos\frac{\theta}{2}=\pm \sqrt{\frac{{1+\cos\theta }}{2}}\) (\(\pm\) 号依 \(\displaystyle\frac{\theta}{2}\)在第几象限而定)

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

上述半角公式的证明是根据二倍角公式:\(\cos 2\alpha= 2{\cos^2}\alpha- 1= 1- 2{\sin^2}\alpha\),

令 \(2\alpha=\theta\) 即 \(\displaystyle\alpha=\frac{\theta}{2}\),移项得 \(\displaystyle 2{\cos ^2}\frac{\theta }{2} = 1+\cos\theta ,2{\sin ^2}\frac{\theta }{2} = 1 -\cos \theta\),

再移项及开平方得 \(\displaystyle\sin \frac{\theta }{2}=\pm\sqrt{\frac{{1-\cos\theta }}{2}}\),\(\displaystyle\cos \frac{\theta }{2}=\pm\sqrt{\frac{{1+\cos\theta }}{2}}\),

将两式相除得 \(\displaystyle\tan \frac{\theta }{2} = \frac{{\sin \frac{\theta }{2}}}{{\cos \frac{\theta }{2}}}=\frac{{\pm\sqrt {\frac{{1 -\cos \theta }}{2}} }}{{\pm\sqrt {\frac{{1 + \cos \theta }}{2}} }} =\pm\sqrt {\frac{{1 – \cos \theta }}{{1 + \cos \theta }}}\),

或根据正弦的二倍角公式 \(\sin{2\theta}\),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 \(\displaystyle\tan \frac{\theta }{2} = \frac{{\sin \frac{\theta }{2}}}{{\cos \frac{\theta }{2}}} = \frac{{2{{\sin }^2}\frac{\theta }{2}}}{{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}} = \frac{{1 – \cos \theta }}{{\sin \theta }}\),

将两式用和比定理得 \(\begin{array}{ll}\displaystyle\tan \frac{\theta }{2}&=\displaystyle\frac{{\sin\theta }}{{1+\cos \theta }}=\frac{{1 – \cos \theta }}{{\sin\theta }}\\&=\displaystyle\frac{{\sin \theta+ (1 -\cos \theta )}}{{1+ \cos\theta+ \sin \theta }}= \frac{{1+\sin \theta- \cos \theta }}{{1 +\sin \theta+\cos \theta }}\end{array}\),

当然也可以使用分比定理得 \(\displaystyle\tan \frac{\theta }{2} = \frac{{ – 1 + \sin \theta+ \cos \theta }}{{1 – \sin \theta+ \cos \theta }}\)。

除了代数证明外,图形证法更容易让学生一目了然,如图一:\(\displaystyle\tan \frac{\varphi }{2} = \frac{{\sin \varphi }}{{1 + \cos \varphi }}\)。

半角公式(I)

图一\(~~~\)来自Wikipedia

另一个图形证明,如图二:\(\displaystyle\tan\frac{\alpha }{2}=\frac{{\sin \alpha }}{{1+ \cos \alpha }}\)。

半角公式(I)

图二\(~~~\)来自Wikipedia

半角公式的目的除了能将三角函数降次之外,另一目的是求一些特殊角的三角函数值,如图三:

\(\begin{array}{ll}\displaystyle\sin {11.25^\circ}&=\displaystyle\sqrt {\frac{{1 – \cos {{22.5}^\circ}}}{2}}= \frac{1}{2}\sqrt {2 – 2 \cdot \frac{1}{2}\sqrt {2 + 2\cos {{45}^\circ}} }\\&=\displaystyle \frac{1}{2}\sqrt {2 – \sqrt {2 + 2 \cdot \frac{{\sqrt 2 }}{2}} }= \frac{1}{2}\sqrt {2 – \sqrt {2 + \sqrt 2 }}\end{array}\)

\(\begin{array}{ll}\displaystyle\cos {11.25^\circ}&=\displaystyle\sqrt {\frac{{1 + \cos {{22.5}^\circ}}}{2}}= \frac{1}{2}\sqrt {2 + 2 \cdot \frac{1}{2}\sqrt {2 + 2\cos {{45}^\circ}} }\\&=\displaystyle \frac{1}{2}\sqrt {2 +\sqrt {2 + 2 \cdot \frac{{\sqrt 2 }}{2}} }= \frac{1}{2}\sqrt {2 + \sqrt {2 +\sqrt 2 } }\end{array}\)

半角公式(I)

图三\(~~~\)单位圆内接正十六边形的部分图形

根据上述的 \(\sin {11.25^\circ} =\frac{1}{2}\sqrt {2-\sqrt{2+\sqrt 2}}\) 值,

我们可从几何图形中,求出单位圆中内接正十六边形的周长,

因为 \({360^\circ} \div 16 = {22.5^\circ}\),也就是上图三 \(\angle BAC=22.5^\circ\),

因为 \(\overline {AD}\bot \overline {BC}\) 且 \(\overline {AB}=\overline {AC}= 1\),得 \(\angle CAD = {22.5^\circ} \div 2 = {11.25^\circ}\),

根据正弦的定义得 \(\overline {DC}=\overline {AC}\cdot \sin {11.25^\circ} = \frac{1}{2}\sqrt {2 – \sqrt {2 + \sqrt 2 } }\),

所以,单位圆中内接正十六边形的周长为 \(16 \times 2\overline{DC}= 16\sqrt {2 – \sqrt {2 + \sqrt 2 } }\)。

连结:半角公式(II)


参考资料


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