回復 2# dtc5527 的帖子
5.
\( \displaystyle 4x^2+y^2+2z^2=10 \to 5-z^2=\frac{4x^2+y^2}{2}\ge 2xy \)
\( \to \sqrt{2xy}+3z \le \sqrt{5-z^2}+3z=\sqrt{5(1-cos^2 a)}+3 \sqrt{5}cos a=\sqrt{5}sin a+3 \sqrt{5}cos a \le 5 \sqrt{2} \)
(上式令\( z=\sqrt{5}cos a \)再用三角疊合)
8.
\( \displaystyle a=tan20^{\circ}+4sin20^{\circ}=\frac{sin20^{\circ}+2(2sin20^{\circ}cos20^{\circ})}{cos20^{\circ}}=\frac{sin20^{\circ}+2sin(60^{\circ}-20^{\circ})}{cos20^{\circ}} \)
\( \displaystyle =\frac{sin20^{\circ}+2(\frac{\sqrt{3}}{2}cos20^{\circ}-\frac{1}{2}sin20^{\circ})}{cos20^{\circ}}=\frac{\sqrt{3}cos20^{\circ}}{cos20^{\circ}}=\sqrt{3} \)
\( a^2=3 \)
