三角函数两角和差公式推导

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一.几何推理

1.两角和公式

做一斜边为1的直角△ABC,任意旋转非$k\Pi ,k=N$,补充如图,令$\angle ABC=\angle \alpha ，\angle CBF=\angle \beta$
$\therefore \angle DBF=\angle DBA+\angle \alpha +\angle \beta =90°,\angle DAF=\angle DBA+\angle DAB$
$\because \angle DAB=\angle \alpha +\angle \beta$
$\therefore \angle ACF+\angle BCF=90°$
$\because \angle ACF=\angle \beta$
$\therefore AB长度为1$
$\because AC=sin(\alpha ),BC=cos(\alpha )$
$\because BF=cos(\alpha )\ast cos(\beta ),CF=cos(\alpha )\ast sin(\beta ),AE=sin(\alpha )sin(\beta ),CE=sin(\alpha )cos(\beta ),BD=EF=sin(\alpha +\beta ),DA=cos(\alpha +\beta )$
$\because \{\begin{array}{l}cos(\alpha +\beta )=cos(\alpha )\ast cos(\beta )-sin(\alpha )\ast sin(\beta )\\ sin(\alpha +\beta )=sin(\alpha )\ast cos(\beta )+cos(\alpha )\ast sin(\beta )\end{array}$

2.两角差公式

$\because \{\begin{array}{l}cos(\alpha +\beta )=cos(\alpha )\ast cos(\beta )-sin(\alpha )\ast sin(\beta )\\ sin(\alpha +\beta )=sin(\alpha )\ast cos(\beta )+cos(\alpha )\ast sin(\beta )\end{array}$
$对\angle \beta 做取反变化$
$\because \{\begin{array}{l}cos(\alpha +(-\beta ))=cos(\alpha )\ast cos(\beta )-sin(\alpha )\ast (-sin(\beta ))\\ sin(\alpha +(-\beta ))=sin(\alpha )\ast cos(\beta )+cos(\alpha )\ast (-sin(\beta ))\end{array}$

$\because \{\begin{array}{l}cos(\alpha -\beta )=sin(\alpha )\ast sin(\beta )+cos(\alpha )\ast sin(\beta )\\ sin(\alpha -\beta )=sin(\alpha )\ast cos(\beta )-cos(\alpha )\ast sin(\beta )\end{array}$

3.总结

$\because \{\begin{array}{l}cos(\alpha +\beta )=cos(\alpha )\ast cos(\beta )-sin(\alpha )\ast sin(\beta )\\ sin(\alpha +\beta )=sin(\alpha )\ast cos(\beta )+cos(\alpha )\ast sin(\beta )\\ cos(\alpha -\beta )=sin(\alpha )\ast sin(\beta )+cos(\alpha )\ast sin(\beta )\\ sin(\alpha -\beta )=sin(\alpha )\ast cos(\beta )-cos(\alpha )\ast sin(\beta )\end{array}$

4.其他

为什么几何推理∠β和∠α不是钝角,根据诱导公式可将钝角化为锐角。所以只推导锐角和可以等价于推导任意角和