![](https://lstatic.shangxueba.com/jiandati/h5/images/m_q_title.png)
A.-Aωsinφ
B.Aωsinφ
C.-Aωcosφ
D.Aωcosφ
A.-Aωsinφ
B.Aωsinφ
C.-Aωcosφ
D.Aωcosφ
A.AB×cosθ1+BC×sinθ2=xC AB×sinθ1+BC×cosθ2=-e
B.AB×cosθ1+BC×cosθ2=xC AB×sinθ1+BC×sinθ2=-e
C.AB×sinθ1+BC×cosθ2=xC AB×cosθ1+BC×sinθ2=-e
D.AB×sinθ1+BC×sinθ2=xC AB×cosθ1+BC×cosθ2=-e
A.ParametricPlot[{Cos[t],Cos[2t]},{sin[t],sin[2t]},{t,0,2Pi}]
B.ParametricPlot[{{Cos[t],Cos[2t]},{Sin[t],Sin[2t]}},{t,0,2Pi}]
C.ParametricPlot[[{Cos[t],Cos[2t]},Sin[t],Sin[2t]}],{t,0,2Pi}]
D.ParametricPlot[{{Cos[t],Cos[2t]},{Sin[t],Sin[2t]}},{t,2Pi,0}]
A.ParametricPlot[{Cos[t],Cos[2t]},{sin[t],sin[2t]},{t,0,2Pi}]
B.ParametricPlot[{{Cos[t],Cos[2t]},{Sin[t],Sin[2t]}},{t,0,2Pi}]
C.ParametricPlot[[{Cos[t],Cos[2t]},Sin[t],Sin[2t]}],{t,0,2Pi}]
D.ParametricPlot[{{Cos[t],Cos[2t]},{Sin[t],Sin[2t]}},{t,2Pi,0}]
A、
B、cos(10t-2)
C、sin(10t+20)
D、sin(10t-20)
A.(cosα, sinαcosΦ,sinαsinΦ)
B.(sinαcosΦ,sinαsinΦ,cosα)
C.(sinαsinΦ,sinαcosΦ,cosα)
D.(consαcosΦ,consαsinΦ,cosα)
A.Cancel[(Sin[2x]-Cos[2x]+1)/(Sin[2x]+Cos[2x]+1),Trig→True]
B.TrigFactor[(Sin[2x]-Cos[2x]+1)/(Sin[2x]+Cos[2x]+1)]
C.Simplify[(Sin[2x]-Cos[2x]+1)/(Sin[2x]+Cos[2x]+1),Trig→True]
D.TrigReduce[(Sin[2x]-Cos[2x]+1)/(Sin[2x]+Cos[2x]+1)]
A.Cancel[(Sin[2x]-Cos[2x]+1)/(Sin[2x]+Cos[2x]+1),Trig→True]
B.TrigFactor[(Sin[2x]-Cos[2x]+1)/(Sin[2x]+Cos[2x]+1)]
C.Simplify[(Sin[2x]-Cos[2x]+1)/(Sin[2x]+Cos[2x]+1),Trig→True]
D.TrigReduce[(Sin[2x]-Cos[2x]+1)/(Sin[2x]+Cos[2x]+1)]
A.AC+AB×cosθ1=BC×cosθ3 AB×cosθ1=BCcos×θ3
B.AC+AB×cosθ1=BC×cosθ3 AB×sinθ1=BC×sinθ3
C.AB×cosθ1=BC×cosθ3 AC+AB×sinθ1=BC×sinθ3
D.AB×sinθ1=BC×cosθ3 AC+AB×cosθ1=BC×sinθ3
为了保护您的账号安全,请在“简答题”公众号进行验证,点击“官网服务”-“账号验证”后输入验证码“”完成验证,验证成功后方可继续查看答案!