1)tan(45°+a)=cosa+sina/cosa-sina
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1)tan(45°+a)=cosa+sina/cosa-sina
2)tan(x+y)tan(x-y)=tan^2x-tan^2y/1-tan^2xtan^y
3)tanx+tany/tanx-tany=sin(x+y)/sin(x-y)
2)tan(x+y)tan(x-y)=tan^2x-tan^2y/1-tan^2xtan^y
3)tanx+tany/tanx-tany=sin(x+y)/sin(x-y)
![1)tan(45°+a)=cosa+sina/cosa-sina](/uploads/image/z/11892472-16-2.jpg?t=1%29tan%2845%C2%B0%2Ba%EF%BC%89%3Dcosa%2Bsina%2Fcosa-sina)
1
tan(45°+a)=1+tana/1-tana=cosa+sina/cosa-sina
2
tan(x+y)tan(x-y)=[(tanx+tany)(tanx-tany)]/[(1-tanxtany)(1+tanxytany)]
=(tan²x-tan²y)/(1-tan²xtan²y)
3.
tanx+tany/tanx-tany
=[(sinx/cosx)+(siny/cosy)]/[(sinx/cosx)-(siny/cosy)]
=(sinxcosy+sinycosx)/(sinxcosy-cosxsiny)
=sin(x+y)/sin(x-y)
tan(45°+a)=1+tana/1-tana=cosa+sina/cosa-sina
2
tan(x+y)tan(x-y)=[(tanx+tany)(tanx-tany)]/[(1-tanxtany)(1+tanxytany)]
=(tan²x-tan²y)/(1-tan²xtan²y)
3.
tanx+tany/tanx-tany
=[(sinx/cosx)+(siny/cosy)]/[(sinx/cosx)-(siny/cosy)]
=(sinxcosy+sinycosx)/(sinxcosy-cosxsiny)
=sin(x+y)/sin(x-y)
1)tan(45°+a)=cosa+sina/cosa-sina
证明(1-cos^2a)/(sina+cosa)-(sina+cosa)/(tan^2a-1)=sina+cosa
证明(1-cos^2a)/(sina-cosa)-(sina+cosa)/(tan^2-1)=sina+cosa
求证:1-cos^2a/sina-cosa - sina+cosa/tan^2a-1=sina+cosa
化简sina/(1-cosa)×根号(tan a-sina/tan a+sina)
求证,cos²a-sina×cosa+tana/cos²a+sina×cosa-tana=1+tan
tan(a/2)=sina/(1+cosa)=(1-cosa)/sina
tan a/2=3 1-cosa-sina/1+cosa+sina
证明(1+sinA-cosA)/(1+sinA+cosA)=tan(A/2)
(cosa-sina)/(cosa+sina)=(1-tana)/(1+tana)=tan(π/4-a)
已知tan(π-a)=2分之1,则2sina-cosa分之sina+cosa=
tan(a/2)=sina/(1+cosa) 怎样证明