作业帮三倍角公式推导,举个例子：sin3a=sin(2a+a)=sin2acosa+cos2asina=2sina(1-sin²a)+(1-2sin²a)sina=3sina-4sin³asin2acosa+cos2asina=2sina(1-sin²a)+(1-2sin²a)sina就这两步之间怎么推？
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/07 08:44:51
三倍角公式推导,举个例子：sin3a=sin(2a+a)=sin2acosa+cos2asina=2sina(1-sin²a)+(1-2sin²a)sina=3sina-4sin³asin2acosa+cos2asina=2sina(1-sin²a)+(1-2sin²a)sina就这两步之间怎么推？
三倍角公式推导,举个例子：
sin3a
=sin(2a+a)
=sin2acosa+cos2asina
=2sina(1-sin²a)+(1-2sin²a)sina
=3sina-4sin³a
sin2acosa+cos2asina
=2sina(1-sin²a)+(1-2sin²a)sina
就这两步之间怎么推？
三倍角公式推导,举个例子：sin3a=sin(2a+a)=sin2acosa+cos2asina=2sina(1-sin²a)+(1-2sin²a)sina=3sina-4sin³asin2acosa+cos2asina=2sina(1-sin²a)+(1-2sin²a)sina就这两步之间怎么推？
sin2acosa+cos2asina
=2sinacosa*cosa+(1-2sin²a)*sina (sin2a=2sinacosa cos2a=1-2sin²a)
=2sina(1-sin²a)+(1-2sin²a)sina
=3sina-4sin³a
三倍角公式:
sin(3α) = 3sinα-4sin³α = 4sinα•sin(60°+α)sin(60°-α)
cos(3α) = 4cos³α-3cosα = 4cosα•cos(60°+α)cos(60°-α)
tan(3α) = (3tanα-tan³α)/(1-3tan²α) = ...
全部展开
三倍角公式:
sin(3α) = 3sinα-4sin³α = 4sinα•sin(60°+α)sin(60°-α)
cos(3α) = 4cos³α-3cosα = 4cosα•cos(60°+α)cos(60°-α)
tan(3α) = (3tanα-tan³α)/(1-3tan²α) = tanαtan(60°+α)tan(60°-α)
三倍角公式推导:
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(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)
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)
收起