Question

已知sin(x-3派/4)cos(x-派/4)=-1/4，求cos4x的值

过程

Answer 1

已知sin(x-3π/4)cos(x-π/4)=-1/4，求cos4x的值

解：因为sin(x-3π/4)=sin[x-(π-π/4)]=sin[-π+(x-π/4)]=sin{-[π-(x-π/4)]}=-sin[π-(x-π/4)]=-sin(x-π/4)，
故原式可化为 -sin(x-π/4)cos(x-π/4)=-1/4；
即有sin(x-π/4)cos(x-π/4)=1/4；
于是得(1/2)sin[2(x-π/4)]=(1/2)sin(2x-π/2)=(1/2)sin[-(π/2-2x)]= -(1/2)sin(π/2-2x)= -(1/2)cos2x=1/4
故得cos2x= -1/2；cos4x=2cos²2x -1=2(1/4) -1=1/2-1= -1/2

Answer 2

sin(x－3π/4)cos(x－π/4)=－1/4
sin(x－π/4)cos(x－π/4)=－1/4
2sin(x－π/4)cos(x－π/4)=－1/2
sin[2(x－π/4)]=－1/2
sin(2x－π/2)=－1/2
－sin(π/2－2x)=－1/2
sin(π/2－2x)=1/2
cos2x=1/2
则：
cos4x=2cos²2x－1=－1/2

Answer 3

sin(x-3π/4)cos(x-π/4)=sin(π-（-x+π/4）)cos(π-（-x+π/4）)=sin(-x+π/4)cos（-x+π/4）
=1/2sin(-2x+π/2)=-1/2cos2x=-1/4
cos2x=1/2
cos4x=2cos²2x-1=1/2-1=-1/2

Answer 4

f(x)=2(sinx)^4 2(cosx)^4 (cos2x)^2-3
=2*((1-cos2x)/2)^2 2*((1 cos2x)/2)^2 (cos2x)^2-3
=(1-cos2x)^2/2 (1 cos2x)^2/2 (cos2x)^2-3
=(cos2x)^2 (cos2x)^2-3
=2(cos2x)^2-3
=2(1 cos4x)/2-3
=cos4x-2
周期=π/2
π/16=〈X〈=3π/16，π/4=〈4X〈=3π/4
-2/2=〈cos4x〈=√2/2
则最小值是√2/2-2