三角形中一个恒等式的证明

Wed, 10 Sep 2025 10:02:37 +0800

定理

三角形$\triangle ABC$中，设角$A$、$B$、$C$所对的边分别为$a$、$b$、$c$。于是有

$$ \frac{\cos A}{\sin B \sin C} + \frac{\cos B}{\sin A \sin C} + \frac{\cos C}{\sin B \sin A} = 2 $$

左式通分，得到

$$ LHS = \frac{\cos A \sin A + \cos B \sin B + \cos C \sin C}{\sin A \sin B \sin C} $$

上下同乘$2$

$$ = \frac{2 \cos A \sin A + 2 \cos B \sin B + 2 \cos C \sin C}{2 \sin A \sin B \sin C} $$

利用二倍角公式，将前两项转换

$$ = \frac{\sin 2A + \sin 2B + 2 \sin C \cos C}{2 \sin A \sin B \sin C} $$

然后和差化积

$$ = \frac{2 \sin(A+B) \cos(A-B) + 2 \cos C \sin C}{2 \sin A \sin B \sin C} $$

$$ \because A + B + C = \pi, \therefore \sin(A+B) = \sin(\pi - C) = \sin C $$

$$ LHS = \frac{2 \sin C \cos(A-B) + 2 \cos C \sin C}{2 \sin A \sin B \sin C} $$

$$ = \frac{\sout {2 \sin C} \cos(A-B) + \sout{2} \cos C \sout{\sin C}}{\sout{2} \sin A \sin B \sout{\sin C}} $$

$$ = \frac{\cos(A-B) + \cos C}{\sin A \sin B} $$

然后积化和差

$$ = \cfrac{\cos(A-B) + \cos C}{-\cfrac{1}{2}[\cos(A+B) - \cos(A-B)]} $$

$$ \because \cos(A+B) = \cos(\pi - C) = - \cos C $$

$$ \therefore LHS = \cfrac{\cos(A-B) + \cos C}{-\cfrac{1}{2}[-\cos C - \cos(A-B)]} $$

$$ = 2\frac{\cos(A-B) + \cos C}{\cos C + \cos(A-B)} $$

$$ = 2 = RHS $$

$\square$