Question

求函数y=(sinx+根号2)*(cosx+根号2)的值域为

Answer 1

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|