三倍角公式推导啊

1个回答

  • sin3a

    =sin(2a+a)

    =sin2acosa+cos2asina

    =2sina(1-sin²a)+(1-2sin²a)sina

    =3sina-4sin³a

    cos3a

    =cos(2a+a)

    =cos2acosa-sin2asina

    =(2cos²a-1)cosa-2(1-cos²a)cosa

    =4cos³a-3cosa

    sin3a=3sina-4sin³a

    =4sina(3/4-sin²a)

    =4sina[(√3/2)²-sin²a]

    =4sina(sin²60°-sin²a)

    =4sina(sin60°+sina)(sin60°-sina)

    =4sina*2sin[(60+a)/2]cos[(60°-a)/2]*2sin[(60°-a)/2]cos[(60°-a)/2]

    =4sinasin(60°+a)sin(60°-a)

    cos3a=4cos³a-3cosa

    =4cosa(cos²a-3/4)

    =4cosa[cos²a-(√3/2)²]

    =4cosa(cos²a-cos²30°)

    =4cosa(cosa+cos30°)(cosa-cos30°)

    =4cosa*2cos[(a+30°)/2]cos[(a-30°)/2]*{-2sin[(a+30°)/2]sin[(a-30°)/2]}

    =-4cosasin(a+30°)sin(a-30°)

    =-4cosasin[90°-(60°-a)]sin[-90°+(60°+a)]

    =-4cosacos(60°-a)[-cos(60°+a)]

    =4cosacos(60°-a)cos(60°+a)

    上述两式相比可得

    tan3a=tanatan(60°-a)tan(60°+a)