cos²x+sinxcosx+1
=（1+cos2x)/2+(sin2x)/2+1
=(sin2x+cos2x)/2+3/2
=√2/2×sin(2x+π/4)+3/2
(1)當(dāng)sin(2x+π/4)=1時(shí),y取得最大值ymax=√2/2+3/2,
此時(shí)2x+π/4=2kπ+π/2 ,x=kπ+π/8,{x|x=kπ+π/8,k∈Z}
(2)y=cos²x+sinxcosx+1=√2/2×sin(2x+π/4)+3/2=√2/2×sin2(x+π/8)+3/2；
y=sinx的圖像①x縮小為原來(lái)的1/2倍得y=sin2x;
②x軸方向左移π/8個(gè)單位得y=sin2（x+π/8）=sin(2x+π/4);
③y縮小為原來(lái)的√2/2倍得y=√2/2×sin(2x+π/4);
④y軸方向上移3/2個(gè)單位得y=√2/2×sin(2x+π/4)+3/2即y=cos²x+sinxcosx+1