已知f（x）=a（sinx+cosx）-sin2x（a∈R）求f（x）在【0，π/2】上的最大值g（a）的表达式
解析：当a=0时，f(x)=-sin2x
极大值点：2x=2kπ-π/2==>x=kπ-π/4；极小值点：2x=2kπ+π/2==>x=kπ+π/4；
当a≠0时，f(x)=√2asin(x+π/4)-sin2x
令f’(x)=√2acos(x+π/4)-2cos2x=√2acos(x+π/4)-2sin(2x+π/2)
=√2acos(x+π/4)-4sin(x+π/4)cos(x+π/4)
=[√2a-4sin(x+π/4)]cos(x+π/4)=0
cos(x+π/4)=0==>x+π/4=2kπ+π/2==>x=2kπ+π/4；x+π/4=2kπ-π/2==>x=2kπ-3π/4
√2a-4sin(x+π/4)=0==> sin(x+π/4)=√2a/4
设θ=arcsin(√2a/4)(-2√2<=a<2√2)
x+π/4=2kπ+θ==>x=2kπ+θ-π/4；x+π/4=2kπ+π-θ==>x=2kπ+3π/4-θ
∴x1=2kπ-3π/4，x2=2kπ+θ-π/4，x3=2kπ+π/4，x4=2kπ+3π/4-θ
f’’(x)=-√2asin(x+π/4)+4sin2x
f’’(x1)>0，f’’(x2)<0，f’’(x3)>0，f’’(x1)<0
∴f(x)在x1=2kπ-3π/4，x3=2kπ+π/4处取极小值；在x2=2kπ+θ-π/4，x4=2kπ+3π/4-θ处取极大值；
∵区间【0，π/2】
当a=0时，f(x)在区间【0，π/2】上最大值为f(0)=f(π/2)=0
当a<2√2时, f(x)在区间【0，π/2】上最大值为f(0)=f(π/2)=a
当a>=2√2时, f(x)在区间【0，π/2】上最大值为f(π/4)=√2a-1
下图中绘出a=-1时f(x)曲线（绿色）；a=1时f(x)曲线（红色）；a=2√2时f(x)曲线（黒色）；
令t=sinx+cosx属于【2,2根号2】
下面就好办了~~