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

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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
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

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
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

①sin²x+cos²x=1
两边平方 则
(sin^4 x+cos^4 x)+2(sin²xcos²x)=1
则 sin²xcos²x=3/16
cos4x=cos²2x-sin²2x
=(cos²x-sin²x)²-4sin²xcos...

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①sin²x+cos²x=1
两边平方 则
(sin^4 x+cos^4 x)+2(sin²xcos²x)=1
则 sin²xcos²x=3/16
cos4x=cos²2x-sin²2x
=(cos²x-sin²x)²-4sin²xcos²x
=sin^4x+cos^4x - 6sin²xcos²x
=5/8-9/8 = -1/2
② sin^4x-cos^4x=
(sin²x+cos²x)(sin²x-cos²x)=sin²x-cos²x=-4/5
则 cos2x = cos²x-sin²x =4/5
所以 sin2x=3/5 或者 -3/5

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