回復 11# ibvtys 的帖子
簡答第3題
已知\(A\)、\(B\)、\(C\)為\(\Delta ABC\)的內角,若\(\displaystyle z=\frac{\sqrt{65}}{5}sin\frac{A+B}{2}+i cos\frac{A-B}{2}\),且\(\displaystyle |\;z|\;=\frac{3\sqrt{5}}{5}\),則\(tan(A+B)\)的最小值為何?
[解答]
\(\begin{align}
& {{\left| z \right|}^{2}}=\frac{13}{5}{{\sin }^{2}}\left( \frac{A+B}{2} \right)+{{\cos }^{2}}\left( \frac{A-B}{2} \right)={{\left( \frac{3}{5}\sqrt{5} \right)}^{2}}=\frac{9}{5} \\
& 13\left( \frac{1-\cos \left( A+B \right)}{2} \right)+5\left( \frac{1+\cos \left( A-B \right)}{2} \right)=9 \\
& 5\cos \left( A-B \right)=13\cos \left( A+B \right) \\
& 5\left( \cos A\cos B+\sin A\sin B \right)=13\left( \cos A\cos B-\sin A\sin B \right) \\
& 18\sin A\sin B=8\cos A\cos B \\
& \tan A\tan B=\frac{4}{9} \\
& \\
& \tan A=x,\tan B=\frac{4}{9x} \\
& \tan \left( A+B \right)=\frac{x+\frac{4}{9x}}{1-\frac{4}{9}}\ge \frac{\frac{4}{3}}{\frac{5}{9}}=\frac{12}{5} \\
\end{align}\)