sin（x－3兀／4）cos（x－兀／4）＝－1／4,求cos4x

原式=[sinx*cos(3π/4) - cosx*sin(3π/4)]*[cosx*cos(π/4) + sinx*sin(π/4)]
=[(-√2/2)sinx - (√2/2)cosx]*[(√2/2)cosx + (√2/2)sinx]
=-[(√2/2)sinx + (√2/2)cosx]*[(√2/2)cosx + (√2/2)sinx]
=-(√2/2)^2*(cosx + sinx)^2
=-(1/2)[(cosx)^2 + (sinx)^2 + 2ainxcosx]
=-(1/2)(1 + sin2x) = -1/4
则：1 + sin2x = 1/2
sin2x=-1/2
cos4x =1 - 2*(sin2x)^2 =1 - 2*(-1/2)^2 =1 - (1/2) =1/2

sin（x－3兀／4）cos（x－兀／4）＝－1／4
sin(x+兀/4-兀)cos（x－兀／4）＝－1／4
-sin(x+兀/4)cos（x－兀／4）＝－1／4
sin(x+兀/4)cos（x－兀／4）＝1／4
(sinxcos兀／4+sin兀／4cosx)(cosxcos兀／4+sinxsin兀／4)=1/4
√2/2(sinx+cosx)*√2/2(sinx+cosx)=1/4
1/2(sinx+cosx)^2=1/4
(sinx+cosx)^2=1/2
1+2sinxcosx=1/2
sin2x=-1/2 (2sinxcosx=sin2x)
(sin2x)^2=1/4
所以cos4x=1-2(sin2x)^2=1-2*(1/4)=1/2

sin(x－3兀/4)cos(x－兀/4)＝(1/2){sin[(x－3兀/4)+(x－兀/4)]+sin[(x－3兀/4)-(x－兀/4)]}=-1/4
sin(2x-兀)+sin(-兀/2)=-1/2,-sin(兀-2x)-1=-1/2,sin2x=-1/2,cos4x=1-2(sin2x)^2=1-2*(1/4)=1/2

化简出来 (sinx+cosx)^2=1/2 1+2sinxcosx=1/2 sin2x=-1/2 cos4x=1-2(sin2x)^2=1/2

sin(x－3π/4) cos(x - π/4) = - 1/4,求cos4x 将要用到：2sinAcosB = sin(A+B) + sin(A-B) 和 cos2α = cos²α - sin²α = 1 - 2sin²α sin(x－3π/4) cos(x - π/4) = 1/2 [sin(2x -π) + sin(-π/...