推 david7834:這種解法也很好^^ 61.225.194.65 01/16 12:27
※ 引述《bitchdog (超人會隱形)》之銘言：
: Use any method you like to find the local minimum and
: local maximum of the following functin :
: f(t)= sint
: ------ , t<2π
: 2+cost
用半角代換做吧!
令u=tan(t/2),t=2arctanu,-π <= t <= π
則sint=2u/(1+u^2),cost=(1-u^2)/(1+u^2)
所以f(u)=[2u/(1+u^2)]/{2+[(1-u^2)/(1+u^2)]}
=2u/(3+u^2)
=>f'(u)=2(3-u^2)/(3+u^2)^2=0
解得u=sqrt3 or -sqrt3
剛好可配一個local max.與另一個local min.
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