2个回答
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y=sinx/2(sin x/2-cos x/2)
=sin²x/2-sinx/2*cos π/2
=½(1-cosx)-½sinx
=-½(sinx+cosx)+1/2
=-√2/2sin(x+π/4)+1/2
当sin(x+π/4)=1时,y=sinx/2(sin x/2-cos x/2)有最小值:(-√2+1)/2
当sin(x+π/4)=-1时,y=sinx/2(sin x/2-cos x/2)有最大值:(√2+1)/2
题目应该是:y=sinx/2(sin x/2-cos x/2)吧
=sin²x/2-sinx/2*cos π/2
=½(1-cosx)-½sinx
=-½(sinx+cosx)+1/2
=-√2/2sin(x+π/4)+1/2
当sin(x+π/4)=1时,y=sinx/2(sin x/2-cos x/2)有最小值:(-√2+1)/2
当sin(x+π/4)=-1时,y=sinx/2(sin x/2-cos x/2)有最大值:(√2+1)/2
题目应该是:y=sinx/2(sin x/2-cos x/2)吧
追问
也许是我抄错了吧
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