关于三角恒等式1.已知cos(α-β)=1/4,cos(α+β)=2/3,求tanαtanβ2.已知∏/2

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  • cos(a-b)=cosacosb+sinasinb=1/4

    cos(a+b)=cosacosb-sinasinb=2/3

    所以sinasinb=-5/24 cosacosb=11/24

    tanatanb=sinasinb/cosacosb=-5/24/(11/24)=-5/11

    -π/4<a-b<0 π<a+b<3π/2

    所以sin(a-b)=-√(1-cos^2(a-b)=-5/13

    cos(a+b)=-√(1-sin^2(a+b)=-4/5

    cos2a=cos[(a-b)+(a+b)]=cos(a-b)cos(a+b)-sin(a-b)sin(a+b)

    =12/13*(-4/5)+5/13*(-3/5)=-63/65

    sin2b=sin[(a+b)-(a-b)]=sin(a+b)cos(a-b)-cos(a+b)sin(a-b)

    =-3/5*12/13-4/5*5/13=-56/65