由 galaxylee 於 星期日 三月 19, 2006 1:03 pm
#ed_op#DIV#ed_cl#1.微分法#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#2.算幾不等式#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#考慮(sinθ)^4*(cosθ)^2的範圍#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#由(sinθ)^2+(cosθ)^2=1 得 (1/2)(sinθ)^2+(1/2)(sinθ)^2+(cosθ)^2=1#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#(1/2)(sinθ)^2+(1/2)(sinθ)^2+(cosθ)^2≧3[(1/4)(sinθ)^4*(cosθ)^2]^(1/3)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#=> (sinθ)^4*(cosθ)^2≦4/27#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#=>-2√3/9≦(sinθ)^2*(cosθ)≦2√3/9#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#等號成立於(1/2)(sinθ)^2=(cosθ)^2=1/3#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#即cosθ=√3/3,θ= ±cos^(-1)√3/3 +2kπ時,最大值2√3/9#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl#