f(x)=sinx(1-cosx)最大值及方法(自變量還是叫x方便一點(diǎn))f(x)=sinx(1-cosx)=sinx-sinxcosx=sinx-(1/2)sin2x令y₁=sinx,y₂=- (1/2)sin2x；y₁的周期是2π,y₂的周期是π；疊加以后y₁+yS...抱歉，這好像不是正確答案。不對(duì)嗎？我用另一種方法求解一下看看：f(x)=sinx-(1/2)sin2x；令f′(x)=cosx-cos2x=cosx-(2cos²x-1)=-2cos²x+cosx+1=-(2cos²x-cosx-1)=-(2cosx+1)(cosx-1)=0，得cosx=-1/2，cosx=1；由cosx=-1/2在[0，2π]內(nèi)得駐點(diǎn)x₁=π-π/3=2π/3；x₂=π+π/3=4π/3；x₁是極大點(diǎn)，x₂是極小點(diǎn)。故極大值f(x)=f(2π/3)=sin(2π/3)-(1/2)sin(4π/3)=sin(π/3)+(1/2)sin(π/3)=(3/2)sin(π/3)=(3/4)√3極小值f(x)=f(4π/3)=sin(4π/3)-(1/2)sin(8π/3)=-sin(π/3)-(1/2)sin(π/3)=-(3/2)sin(π/3)=-(3/4)√3.由cosx=0在[0，2π]內(nèi)得駐點(diǎn)x₃=π/2；x₄=3π/2；x₃是極大點(diǎn)，x₄.是極小點(diǎn)。極大值f(x)=f(π/2)=sin(π/2)-(1/2)sinπ=1；極小值f(x)=f(3π/2)=sin(3π/2)-(1/2)sin(3π)=-1.故f(x)的最大值是(3/4)√3，最小值是-(3/4)√3。前面的回答真是錯(cuò)的！對(duì)不起啦，沒(méi)有仔細(xì)檢查。