已知sin(x-135)cos(x-45)=-1/4,求cos4x

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已知sin(x-135)cos(x-45)=-1/4,求cos4x
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已知sin(x-135)cos(x-45)=-1/4,求cos4x
已知sin(x-135)cos(x-45)=-1/4,求cos4x

已知sin(x-135)cos(x-45)=-1/4,求cos4x
sin(x-135)cos(x-45)= 1/2 [ (cosx)^2 -(sinx)^2 ] = -1/4; => (cosx)^2 -(sinx)^2=-1;
又因为 (cosx)^2 +(sinx)^2= 1; 综合以上可以解出 (cosx)^2=0,(sinx)^2=1;
所以cos4x = cos (2x+2x) = (cos2x)^2 -(sin2x)^2
= [(cosx)^2 -(sinx)^2 ] [(cosx)^2 -(sinx)^2 ] - 4 (sinx)^2 (cosx)^2
= (-1)×(-1) - 4 ×1×0 = 1