cos4x=2cos^2(2x)
(cos2x-1)/2=-sin^2(x)
cos4x-4cos2x+3
=2cos^2(2x)-4cos2x+3
=2(cos^2(2x)-2cos2x+1)
=2(cos2x-1)^2
=8[(cos2x-1)/2]^2
=8[-sin^2(x)]^2
=8sin^4(x)
化简cos4x-4cos2x+3
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