函数y=sin（π／2+x）*（sinx+cosx）的最小值,求步骤

y = sin（π／2+x）*（sinx+cosx）
= cosx *（sinx+cosx）
= sinxcosx+cos²x
= 1/2sin2x + 1/2(cos2x-1)
= 1/2(sin2x+cos2x) - 1/2
= √2/2(sin2xcosπ/4+cos2xsinπ/4) - 1/2
= √2/2sin(2x+π/4) - 1/2
sin(2x+π/4)最小值-1
∴ymin = -√2/2-1/2 = -(√2+1)/2