网刊2期三角公式一文内某段的latex代码直接贴于此看看效果如何
两角和差展开公式:
\begin{align*}
\sin (\alpha \pm \beta ) & = \sin \alpha \cos \beta \pm \cos \alpha \sin \beta &
\cos (\alpha \pm \beta ) & = \cos \alpha \cos \beta \mp \sin \alpha \sin \beta \\
\tan (\alpha \pm \beta ) & = \frac{{\tan \alpha \pm \tan \beta }}{{1 \mp \tan \alpha \tan \beta }} &
\cot (\alpha \pm \beta ) & = - \frac{{1 \mp \cot \alpha \cot \beta }}{{\cot \alpha \pm \cot \beta }}
\end{align*}
由正余弦的和差展开式可得如下的“积化和差”与“和差化积”公式:
\begin{align*}
\sin \alpha \cos \beta & = \frac{1}{2}\left( {\sin \left( {\alpha + \beta } \right) + \sin \left( {\alpha - \beta } \right)} \right) & \sin \alpha + \sin \beta & = 2\sin \frac{{\alpha + \beta }}{2}\cos \frac{{\alpha - \beta }}{2}\\
\cos \alpha \sin \beta & = \frac{1}{2}\left( {\sin \left( {\alpha + \beta } \right) - \sin \left( {\alpha - \beta } \right)} \right) & \sin \alpha - \sin \beta & = 2\cos \frac{{\alpha + \beta }}{2}\sin \frac{{\alpha - \beta }}{2}\\
\cos \alpha \cos \beta & = \frac{1}{2}\left( {\cos \left( {\alpha + \beta } \right) + \cos \left( {\alpha - \beta } \right)} \right) & \cos \alpha + \cos \beta & = 2\cos \frac{{\alpha + \beta }}{2}\cos \frac{{\alpha - \beta }}{2}\\
\sin \alpha \sin \beta & = - \frac{1}{2}\left( {\cos \left( {\alpha + \beta } \right) - \cos \left( {\alpha - \beta } \right)} \right) & \cos \alpha - \cos \beta & = - 2\sin \frac{{\alpha + \beta }}{2}\sin \frac{{\alpha - \beta }}{2}
\end{align*}
以上的公式属于书本上的公式,故此就不写推导了。“积化和差”和“和差化积”公式虽然据说只要求了解不要求记忆,不过我觉得还是不妨一记,因为挺有用的,至少本文中多次用到。另外扯一题外话,不要总以为学得少或不用记就是减负,其实有时学多点记多点有价值的定理和方法,熟悉运用之后不但不是负担,更会成为你解决问题的兵器,所以换个角度来看,适当学多点也可以是减负。
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