1)sin^4x+cos^4x=5/8 求cos4x 2)sin^4x-cos^4x=-4/5求sin2x
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1)sin^4x+cos^4x=5/8 求cos4x 2)sin^4x-cos^4x=-4/5求sin2x
要有过程越详细越好
要有过程越详细越好
1.sin^4x + cos^4x=(sin^2x+cos^2x)^2 - 2sin^2xcos^2x=5/8
∵sin^2x+cos^2x=1
∴1-2sin^2xcos^2x=5/8
2sin^2xcos^2x=3/8
则:(sin2x)^2=3/4
cos4x=1-2(sin2x)^2=1 - 3/2 =-1/2
2.∵sin^4x-cos^4x=(sin^2x+cos^2x)(sin^2x-cos^2x)=-4/5
∴sin^2x -cos^2x=-4/5
cos^2x-sin^2x=4/5=cos2x
sin2x=±3/5
∵sin^2x+cos^2x=1
∴1-2sin^2xcos^2x=5/8
2sin^2xcos^2x=3/8
则:(sin2x)^2=3/4
cos4x=1-2(sin2x)^2=1 - 3/2 =-1/2
2.∵sin^4x-cos^4x=(sin^2x+cos^2x)(sin^2x-cos^2x)=-4/5
∴sin^2x -cos^2x=-4/5
cos^2x-sin^2x=4/5=cos2x
sin2x=±3/5
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