引用:
已知 \(\triangle ABC\) 的三內角為 \(A,B,C\), 求 \(\frac{\cos A}{\sin B\sin C}+\frac{\cos B}{\sin C\sin A}+\frac{\cos C}{\sin A \sin B}\) 之值.
設三角形的外接圓半徑為 \(R\)，面積為 \(S\)，則
\[\frac{\cos A}{\sin B\sin C}+\frac{\cos B}{\sin C\sin A}+\frac{\cos C}{\sin A \sin B}\]
\[=\frac{{\frac{{b^2 + c^2 - a^2 }}{{2bc}}}}{{\frac{b}{{2R}} \cdot \frac{c}{{2R}}}} + \frac{{\frac{{a^2 + c^2 - b^2 }}{{2ac}}}}{{\frac{a}{{2R}} \cdot \frac{c}{{2R}}}} + \frac{{\frac{{a^2 + b^2 - c^2 }}{{2ab}}}}{{\frac{a}{{2R}} \cdot \frac{b}{{2R}}}}\]
\[ = 2R^2 \left( {\frac{{b^2 + c^2 - a^2 }}{{b^2 c^2 }} + \frac{{a^2 + c^2 - b^2 }}{{a^2 c^2 }} + \frac{{a^2 + b^2 - c^2 }}{{a^2 b^2 }}} \right)\]
\[= \frac{{2R^2 }}{{a^2 b^2 c^2 }}\left( {a^2 \left( {b^2 + c^2 - a^2 } \right) + b^2 \left( {a^2 + c^2 - b^2 } \right) + c^2 \left( {a^2 + b^2 - c^2 } \right)} \right)\]
\[= \frac{{2R^2 }}{{a^2 b^2 c^2 }}\left( {2a^2 b^2 + 2b^2 c^2 + 2a^2 c^2 - a^4 - b^4 - c^4 } \right)\]
\[= \left( {\frac{{4R}}{{abc}}} \right)^2 \cdot \frac{1}{{16}}\left[ {\left( {a + b + c} \right)\left( { - a + b + c} \right)\left( {a - b + c} \right)\left( {a + b - c} \right)} \right] \cdot 2\]
\[= \frac{1}{{S^2 }}\left[ {\left( {\frac{{a + b + c}}{2}} \right)\left( {\frac{{ - a + b + c}}{2}} \right)\left( {\frac{{a - b + c}}{2}} \right)\left( {\frac{{a + b - c}}{2}} \right)} \right] \cdot 2\]
\[= \frac{1}{{S^2 }} \cdot S^2 \cdot 2 = 2\]
另解，
\[\frac{{\cos A}}{{\sin B\sin C}} + \frac{{\cos B}}{{\sin A\sin C}} + \frac{{\cos C}}{{\sin A\sin B}} \]
\[= \frac{{\cos \left( {\pi - \left( {B + C} \right)} \right)}}{{\sin B\sin C}} + \frac{{\cos \left( {\pi - \left( {B + C} \right)} \right)}}{{\sin A\sin C}} + \frac{{\cos \left( {\pi - \left( {B + C} \right)} \right)}}{{\sin A\sin B}} \]
\[= - \frac{{\cos \left( {B + C} \right)}}{{\sin B\sin C}} - \frac{{\cos \left( {A + C} \right)}}{{\sin A\sin C}} - \frac{{\cos \left( {A + B} \right)}}{{\sin A\sin B}} \]
\[= - \frac{{\cos B\cos C - \sin B\sin C}}{{\sin B\sin C}} - \frac{{\cos A\cos C - \sin A\sin C}}{{\sin A\sin C}} - \frac{{\cos A\cos B - \sin A\sin B}}{{\sin A\sin B}} \]
\[= - \left( {\cot B\cot C - 1} \right) - \left( {\cot A\cot C - 1} \right) - \left( {\cot A\cot B - 1} \right) \]
\[= - \left( {\cot B\cot C + \cot A\cot C + \cot A\cot B} \right) + 3 \]
\[= - 1 + 3 = 2 \]
註：\(\triangle ABC\) 有性質 \(\cot A\cot B + \cot B\cot C + \cot A\cot C=1.\)
晚點繼續想利用『積化和差、和差化積』的解法。
