已知tanx=2,求下列各式(1)2/3sin2x+1/4cos2x(2)2sin2x-sinxcosx+cos2x

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  • (1)2/3sin2x+1/4cos2x

    =[2/3*2sinxcosx+1/4(cos²x-sin²x)]/(sin²x+cos²x) 两边同除以cos²x

    =(4/3tanx+1/4-1/4tan²x)/(tan²x+1)

    ∵tanx=2

    ∴原式=(4/3*2+1/4-1/4*2²)/(2²+1)=23/60

    (2)2sin2x-sinxcosx+cos2x

    =2*2sinxcosx-sinxcosx+cos²x-sin²x

    =(3sinxcosx+cos²x-sin²x)/(sin²x+cos²x) 两边同除以cos²x

    =(3tanx+1-tan²x)/(tan²x+1)

    ∵tanx=2

    ∴原式=(3*2+1-2²)/(2²+1)=3/5