求f(x)=(a+sinx)(a+cosx)的最值
问题描述：
求f(x)=(a+sinx)(a+cosx)的最值
答
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