有木有三倍角公式
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/09 03:37:40
有木有三倍角公式
有木有三倍角公式
有木有三倍角公式
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)
必须有啊
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...
全部展开
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
收起