1)tan(45°+a)=cosa+sina/cosa-sina2)tan(x+y)tan(x-y)=tan^2x-tan^2y/1-tan^2xtan^y3)tanx+tany/tanx-tany=sin(x+y)/sin(x-y)
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1)tan(45°+a)=cosa+sina/cosa-sina2)tan(x+y)tan(x-y)=tan^2x-tan^2y/1-tan^2xtan^y3)tanx+tany/tanx-tany=sin(x+y)/sin(x-y)
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)
1)tan(45°+a)=cosa+sina/cosa-sina2)tan(x+y)tan(x-y)=tan^2x-tan^2y/1-tan^2xtan^y3)tanx+tany/tanx-tany=sin(x+y)/sin(x-y)
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)
求证:sina-cosa+1/sina+cosa-1=tan(a/2+π/4)(sina-cosa+1)/(sina+cosa-1)=tan(a/2+π/4)
求证(1-cosa)/sina=tan(a/2)
求证(1-cosa)/sina=tan(a/2)
1)tan(45°+a)=cosa+sina/cosa-sina2)tan(x+y)tan(x-y)=tan^2x-tan^2y/1-tan^2xtan^y3)tanx+tany/tanx-tany=sin(x+y)/sin(x-y)
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证明(1-cos^2a)/(sina+cosa)-(sina+cosa)/(tan^2a-1)=sina+cosa
求证(1-cos²a/sina-cosa)-(sina+cosa/tan²a-1)=sina+cosa
证明(1-cos^2a)/(sina-cosa)-(sina+cosa)/(tan^2-1)=sina+cosa
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化简sina/(1-cosa)×根号(tan a-sina/tan a+sina)错了。原题是化简sina/(1-cosa)×根号【(tan a-sina)/(tan a+sina)】,
已知tan(a+45)=2,则1+3sina乘cosa-2cos2a=tan(a+派/4)
证明恒等式(1+sina)/cosa=(1+tan二分之a)/(1-tan二分之a)
求证:(1)(1+sin2a)/(cosa+sina)=cosa+sina (2)(1-cos2a)/(1+cos2a)=tan^2a
已知tan(π-a)=2分之1,则2sina-cosa分之sina+cosa=
tan(45+a)=-2 求2sina/sina-cosa
已知tan(π+a)=-1/3,tan(α+β)=[sin2(π/2-a)+4cosa^2]/[10cosa^2-sin2a].1.求tan(α+β)的值.2求tanβ的值.
若sina-cosa=1/2 tan a=
tan(a/2)=sina/(1+cosa) 怎样证明