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isee
发表于 2013-10-21 10:33
本帖最后由 isee 于 2013-10-21 10:45 编辑 笨方法，设 $\tan(\alpha/2)=a$, $\tan(\beta/2)=b$，条件等式和所求的都化为二者，一定能做。 ...
kuing 发表于 2013-10-18 20:12
第一感亦如此，具体过程如下：
题目：
已知$\dfrac {\cos \alpha \cdot \cos {\dfrac \beta 2}}{\cos (\alpha -\dfrac \beta 2)}+\dfrac {\cos \beta \cdot \cos {\dfrac \alpha 2}}{\cos (\beta -\dfrac \alpha 2)}=1$，求$\cos \alpha+\cos \beta$.
解答：
记$\tan \dfrac \alpha 2=a,\tan \dfrac \beta 2=b$，则$\cos\alpha=\dfrac {\cos^2\frac \alpha 2 -\sin^2\frac \alpha 2}{\cos^2\frac \alpha 2 +\sin^2\frac \alpha 2}=\dfrac {1-a^2}{1+a^2}$.
同样的$\cos \beta=\dfrac {1-b^2}{1+b^2}$
\begin{align*}
\dfrac {\cos \alpha \cdot \cos {\dfrac \beta 2}}{\cos (\alpha -\dfrac \beta 2)}&=\dfrac {\cos \alpha \cdot \cos {\dfrac \beta 2}}{\cos \alpha \cdot \cos\dfrac \beta 2+\sin \alpha \cdot \sin\dfrac \beta 2}\\[2ex]
&=\dfrac {1}{1+\tan \alpha \cdot \tan\dfrac \beta 2}\\[2ex]
&=\dfrac {1}{1+\dfrac {2a}{1-a^2} \cdot b}\\[2ex]
&=\dfrac {1-a^2}{1-a^2+2ab}\\[2em]
\dfrac {\cos \beta \cdot \cos {\dfrac \alpha 2}}{\cos (\beta -\dfrac \alpha 2)}&=\dfrac {1-b^2}{1-b^2+2ab}\\[2ex]
\therefore \dfrac {\cos \alpha \cdot \cos {\dfrac \beta 2}}{\cos (\alpha -\dfrac \beta 2)}+\dfrac {\cos \beta \cdot \cos {\dfrac \alpha 2}}{\cos (\beta -\dfrac \alpha 2)}&=\dfrac {1-a^2}{1-a^2+2ab}+\dfrac{1-b^2}{1-b^2+2ab}=1\\[2ex]
\Rightarrow (1-a^2)(1-b^2+2ab)&+(1-b^2)(1-a^2+2ab)\\&=(1-a^2+2ab)(1-b^2+2ab)\\[2ex]
(1-b^2)(1-a^2+2ab)&=2ab(1-b^2+2ab)\\[2ex]
1-a^2-b^2&=3a^2b^2\\[2ex]
a^2+b^2&=1-3a^2b^2\\[2em]
\therefore \cos \alpha+\cos \beta=\dfrac {1-a^2}{1+a^2}+\dfrac {1-b^2}{1+b^2}&=\dfrac {2-2a^2b^2}{1+a^2+b^2+a^2b^2}\\[2ex]
&=\dfrac {2-2a^2b^2}{1+1-3a^2b^2+a^2b^2}\\[2ex]
&=1
\end{align*} |
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kuing
发表于 2013-10-18 20:12
