求值域y=sinx+cosx+sinxcosx
解:令:sinx+cosx=t 化简得:√2sin(x+π/4)=t 则:-√2≤t≤√2 1+2sinxcosx=t² sinxcosx=(t²-1)/2 y=sinx+cosx+sinxcosx =t+(t²-1)/2 =t²/2+t-1/2 =1/2(t²+2t+1)-1 =1/2(t+1)²-1 那么,当t=-1 y最小=-1 t=√2 y最大=1/2(√2+1)²-1=√2-0.5 所以:函数的值域y∈[0,(√2-0.5)]。
设t=sinx+cosx=√2sin(x+π/4),-√2≤t≤√2. sinxcosx=(t^2-1)/2, ∴y=t+[(t^2-1)/2]=(1/2)(t+1/2)^2-1. ∴当t=-1时,最小值y|min=-1; 当t=√2时,最大值y|max=(1+2√2)/2.
答:【1】y=sinx+cosx+sinxcosx =sinx+cosx+[(sinx+cosx)^2-1]/2 =[(sinx+cosx+1)^2]/2 -1. ...详情>>
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