Question

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

Answer 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]