求函数y=cosx(cosx+sinx)的值域

y=cosx(cosx+sinx)
=cos²x+sinxcosx
=(cos2x+1)/2+1/2·sin2x
=1/2·(sin2x+cos2x)+1/2
=1/2·√2(√2/2·sin2x+√2/2·cos2x)+1/2
=1/2·√2(cosπ/4·sin2x+sinπ/4·cos2x)+1/2
=1/2·√2sin(2x+π/4)+1/2
=√2/2·sin(2x+π/4)+1/2
∵sin(2x+π/4)∈【-1,1】
∴1/2-√2/2≤√2/2·sin(2x+π/4)+1/2≤1/2+√2/2
故值域为：【1/2-√2/2,1/2+√2/2】

y=(cosx)^2+sinxcosx=(cos2x+1)/2+sin2x/2=1/2+(sin2x+cos2x)/2=1/2+√2(sin2xcosπ/4+cos2xsinπ/4)/2=1/2+√2sin(2x+π/4)/2，所以函数的最小值为（1-√2）/2，最大值为（1+√2）/2