几个二阶行列式的问题求函数y=|sin(x + pai/4) sin2x|的值域| 1 1 |

∵-sin2x=sin(-2x)=cos(2x+π/2)=cos[2(x+π/4)]=1-2[sin(x+π/4)]^2
∴y=sin(x+π/4)-sin2x=sin(x+π/4)+1-2[sin(x+π/4)]^2
令t=sin(x+π/4),则
y=-2t^2+t+1,t∈[-1,1]
对称轴t=1/4,开口向下
∴ymax=y(t=1/4)=-1/8+1/4+1=9/8
ymin=y(t=-1)=-2-1+1=-2
故函数值域为[-2,9/8]

sin(x+pi/4)-sin(2x)=sin(x+pi/4)+cos(2*(x+pi/4))=sin(t)+cos(2t)
sin(t)+cos(2t)=sin(t)+1-2*(sin(t))^2 sin(t)=1/4时最大"9/8"，最小"-2"