Let sinx+cosx=T, (1)
According to the trigonometric function relationship sinx cosx = [(sinx+cosx) 2-(sinx 2+cosx 2)]/2.
Substituting the formula (1) gives sinx cosx = (t 2- 1)/2.
So y = t+(t 2- 1)/2.
Finishing, y =1/2 (t+1) 2-1.
And sinx+cosx = √ 2sin (x+π/4) ∈ [-√ 2, √ 2].
So y is not monotonous when T[∈-√2, √2].
When T=- 1, y takes the minimum value =- 1.
When T=√2, the maximum value of y is = 1/2+√2.
Range [- 1, 1/2+√2]
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