求函数y=(sinx+根号2)*(cosx+根号2)的值域为
1个回答

令t=sinx+cosx

则t^2=1+2sinxcosx,得sinxcosx=(t^2-1)/2

y=sinxcosx+√2(sinx+cosx)+2

=(t^2-1)/2+√2t+2

=(t^2+2√2t+3)/2

=[(t+√2)^2+1]/2

因为t=sinx+cosx=√2sin(x+π/4),故|t|

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