已知tanx=2(1)求(2/3)sin^2x+(1/4)cos^2x的值（2）求2sin^x-sinxsosx+cos^2x的值

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已知tanx=2(1)求(2/3)sin^2x+(1/4)cos^2x的值（2）求2sin^x-sinxsosx+cos^2x的值
已知tanx=2(1)求(2/3)sin^2x+(1/4)cos^2x的值
（2）求2sin^x-sinxsosx+cos^2x的值

已知tanx=2(1)求(2/3)sin^2x+(1/4)cos^2x的值（2）求2sin^x-sinxsosx+cos^2x的值
tanx=2
平方得
sin²x=4cos^2x
5cos^2x=1 cos^2x=1/5
sin^2x=4/5
sinxcosx=√[sin^2xcos^2x]=2/5
(2/3)sin^2x+(1/4)cos^2x
=(8/12)sin^2x+(3/12)cos^2x
=(3/12)[sin^2x+cos^2x]+(5/12)sin^2x
=(3/12)+(5/12)sin^2x
=3/12+(5/12)*(4/5)
=7/12
2sin^x-sinxsosx+cos^2x
=1+sin²x-sinxcosx
=1+4/5-2/5
=7/5

1、tanx=2 所以有：sinx=2cosx 可得：
sin^2x+cos^2x=1 即：cos^2x=1/5
(2/3)sin^2x+(1/4)cos^2x
=(2/3)4cos^2x+(1/4)cos^2x
=(35/12)X(1/5)
=7/12
2、2sin^2x-sinxsosx+cos^2x
=8cos^2x-2cos^2x...

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1、tanx=2 所以有：sinx=2cosx 可得：
sin^2x+cos^2x=1 即：cos^2x=1/5
(2/3)sin^2x+(1/4)cos^2x
=(2/3)4cos^2x+(1/4)cos^2x
=(35/12)X(1/5)
=7/12
2、2sin^2x-sinxsosx+cos^2x
=8cos^2x-2cos^2x+cos^2x
=7cos^2x
=7/5

收起

2/3sin^2x+1/4cos^2x
=2/3(sin^2x+cos^2x)-5/12cos^2x
=2/3-5/12/(tan^2x+1)
=7/12
2sin^2x-sinxsosx+cos^2x
=(2tan^2x-tanx+1)*cos^2x
=(2tan^2x-tanx+1)/(tan^2x+1)
=7/5

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