求证(2-2cos^6x-2sin^6x)/(3-3cos^4x-3sin^4x)-(1/4)sin^2(2x)-cos
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求证(2-2cos^6x-2sin^6x)/(3-3cos^4x-3sin^4x)-(1/4)sin^2(2x)-cos^4x=sin^2x
因为2-2cos^6x-2sin^6x=2-2(cos²x+sin²x)(cos^4x+sin^4x-cos²xsin²x)
=2-2[(cos²x+sin²x)²-3cos²xsin²x]=6cos²xsin²x,
3-3cos^4x-3sin^4x=3-3[(cos²x+sin²x)²-2cos²xsin²x]=6cos²xsin²x,
所以,(2-2cos^6x-2sin^6x)/(3-3cos^4x-3sin^4x)=1.
故原式左边=1-(1/4)sin²2x-cos^4x=1-(1/4)sin²2x-[(1+cos2x)/2]²
=1-(1/4)sin²2x-(1+2cos2x+cos²2x)/4
=1-(1/4)(sin²2x+cos²2x)-(1+2cos2x)/4
=1-1/4-1/4-(cos2x)/2=(1-cos2x)/2
=sin²x=右边,
所以,原等式成立.
=2-2[(cos²x+sin²x)²-3cos²xsin²x]=6cos²xsin²x,
3-3cos^4x-3sin^4x=3-3[(cos²x+sin²x)²-2cos²xsin²x]=6cos²xsin²x,
所以,(2-2cos^6x-2sin^6x)/(3-3cos^4x-3sin^4x)=1.
故原式左边=1-(1/4)sin²2x-cos^4x=1-(1/4)sin²2x-[(1+cos2x)/2]²
=1-(1/4)sin²2x-(1+2cos2x+cos²2x)/4
=1-(1/4)(sin²2x+cos²2x)-(1+2cos2x)/4
=1-1/4-1/4-(cos2x)/2=(1-cos2x)/2
=sin²x=右边,
所以,原等式成立.
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