f(x)=(a+sinx)(a+cosx) =sinxcosx+a（sinx+cosx）+a^2.设sinx+cosx=t （-√2≤t≤√2）,则sinxcosx=（t^2-1）/2,f(t)=(t^2+2at+2a^2-1)/2,（-√2≤t≤√2）.问题转化为求f(t)的最值 f(t)=1/2 [(t +a )^2] +(a^2-1)/2...可不可以这样：y=(a+sinx)(a+cosx)=sinxcosx+a(sinx+cosx)+a^2=(1/2)sin2x+√2sin(x+π/4)+a^2=-(1/2)cos(2x+π/2)+√2sin(x+π/4)+a^2=-(1/2){1-2[sin(x+π/4)]^2}+√2sin(x+π/4)+a^2=sin(x+π/4)]^2+√2sin(x+π/4)+a^2-1/2令t=sin(x+π/4)则有-1<=t<=1原式就=（t-√2\2)^2+a^2-1,对称轴为t=-√2\2,则取该值时，有最小值a^2-1取t=-1时，则有最大值a^2+√2你好像在恒等变换中少了个ay=(a+sinx)(a+cosx)=sinxcosx+a(sinx+cosx)+a^2=(1/2)sin2x+√2sin(x+π/4)+a^2应为=-(1/2)cos(2x+π/2)+√2asin(x+π/4)+a^2