sin（六分之七π-α）等于三分之一.求cos(三分之二π+2α)
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解决时间 2021-12-31 04:09
- 提问者网友：嘚啵嘚啵
- 2021-12-30 22:03
sin（六分之七π-α）等于三分之一.求cos(三分之二π+2α)
最佳答案
- 五星知识达人网友：旧脸谱
- 2021-12-30 22:39
sin（六分之七π-α）等于三分之一
-sin(π/6-α)=1/3
sin(π/6-α)=-1/3
cos[π/2-(π/6-α)]=-1/3
cos(π/3+α)=-1/3
cos2(π/3+α)=2cos²(π/3+α)-1
cos(2π/3+2α)=2/9-1=-7/9
-sin(π/6-α)=1/3
sin(π/6-α)=-1/3
cos[π/2-(π/6-α)]=-1/3
cos(π/3+α)=-1/3
cos2(π/3+α)=2cos²(π/3+α)-1
cos(2π/3+2α)=2/9-1=-7/9
全部回答
- 1楼网友：毛毛
- 2021-12-31 00:00
sin(7π/6-α)=1/3
sin(π+π/6-α)=1/3
-sin(π/6-α)=1/3
sin(π/6-α)=-1/3
cos(2π/3+2α)
=cos(π-π/3+2α)
=cos[π-(π/3-2α)]
=-cos(π/3-2α)
=-[1-2sin^2(π/6-α)]
=-[1-2*(-1/3)^2]
=-[1-2/9]
=-7/9
- 2楼网友：骨子里都是戏
- 2021-12-30 23:42
已知tan=sin/cos
所以tan=三分之二/三分之一=2
- 3楼网友：从此江山别
- 2021-12-30 23:10
sin(7π/6-α)=1/3
sin(π+π/6-α)=1/3
sin(π/6-α)= -1/3
cos(2π/3+2α)
=cos(π-π/3+2α)
=cos[π-(π/3-2α)]
=-cos(π/3-2α)
=-[1-2sin^2(π/6-α)]
=-[1-2*(-1/3)^2]
=-[1-2/9]
=-7/9
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