已知tanx=2,求下列格式的值

1个回答

  • 1(2sinx+3cosx)/(4sinx-9cosx)

    =(2tanx+3)/(4tanx-9) (分子分母同时除以cosx,sinx/cosx=tanx)

    =(4+3)/(8-9)=-7

    2 (2sin²x-3cos²x)/(4sin²x+9cos²x) (分子分母同时除以cos²x,sin²x/cos²x=tan²x)

    =(2tan²x-3)/(4tan²x+9)

    =(8-3)/(16+9)=1/5

    3 4sin²x-3sinx*cosx-5cos²x

    =(4sin²x-3sinx*cosx-5cos²x)/(sin²x+cos²x) (sin²x+cos²x=1)

    =(4tan²x-3tanx-5)/(tan²x+1)

    =(8-6-5)/(4+1)=-3/5

    4 (sinx-cos^5x)/(3sin³x-cos³x) (sin²x+cos²x)^2=sin⁴x+cos⁴x+2sin²xcos²x=1

    =[sinx(sin⁴x+cos⁴x+2sin²xcos²x)-cos^5x]/[(3sin³x-cos³x)(sin²x+cos²x]

    =[sin^5x+sinxcos⁴x+2sin³xcos²x)-cos^5x]/[(3sin³x-cos³x)(sin²x+cos²x]

    =(tan^5x+tanx+2tan³x-1)/[(3tan³x-1)(tan²x+1)]

    =(32+2+16-1)/[(24-1)(4+1)]=49/115

    5 23/(3sin³xcosx-cos⁴x)

    =23(sin⁴x+cos⁴x+2sin²xcos²x)/(3sin³xcosx-cos⁴x)

    =23(tan⁴x+1+2tan²x)/(3tan³x-1)

    =23(16+1+8)/(24-1)

    =25