有木有三倍角公式
3个回答

三倍角公式:  sin(3α) = 3sinα-4sin^3α = 4sinα·sin(60°+α)sin(60°-α)

cos(3α) = 4cos^3α-3cosα = 4cosα·cos(60°+α)cos(60°-α)

tan(3α) = (3tanα-tan^3α)/(1-3tan^2α) = tanαtan(π/3+α)tan(π/3-α)

编辑本段三倍角公式推导:

1.sin3a

=sin(2a+a)

=sin2acosa+cos2asina

=2sina(1-sin^2a)+(1-2sin^2a)sina

=3sina-4sin^3a

2.cos3a

=cos(2a+a)

=cos2acosa-sin2asina

=(2cos^2a-1)cosa-2(1-cos^2a)cosa

=4cos^3a-3cosa

(1)sin3a=3sina-4sin^3a

=4sina(3/4-sin^2a)

=4sina[(√3/2)^2-sin^2a]

=4sina(sin^260°-sin^2a)

=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)

(2)cos3a=4cos^3a-3cosa

=4cosa(cos^2a-3/4)

=4cosa[cos^2a-(√3/2)^2]

=4cosa(cos^2a-cos^230°)

=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)

综上述两式相比可得

tan3a=tanatan(60°-a)tan(60°+a)