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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