已知sinx+siny=4/5,codx+cosy=3/5求-查字典问答网
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  已知sinx+siny=4/5,codx+cosy=3/5求cos(x-y)计算[sin50+cos40*(1+tan60*tan10)]/cos^220化简(2cos^2a-1)/[2tan*(45-a)*sin^2(45+a)](我帮弧度制换成角度值了)已知cos(a+b)=1/3cos(a-b)=1/6求tana*tanb的值在△ABC中求证tanA

  已知sinx+siny=4/5,codx+cosy=3/5求cos(x-y)

  计算[sin50+cos40*(1+tan60*tan10)]/cos^220

  化简(2cos^2a-1)/[2tan*(45-a)*sin^2(45+a)](我帮弧度制换成角度值了)

  已知cos(a+b)=1/3cos(a-b)=1/6求tana*tanb的值

  在△ABC中求证tanA+tanB+tanC=tanA*tanB*tanC,

  tan(A/2)*tan(B/2)+tan(B/2)*tan(C/2)+tan(C/2)*tan(A/2)=1

  建议把过程写清楚点适合在试卷上用的语言

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2020-06-24 15:26
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秦元璇

  1.两个式子分别平方:(sinx+siny)^2=(sinx)^2+2sinxsiny+(siny)^2=16/25

  (cosx+cosy)^2=(cosx)^2+2cosxcosy+(cosy)^2=9/25

  两式相加:(sinx)^2+(cosx)^2+2sinxsiny+2cosxcosy+(siny)^2+(cosy)^2=16/25+9/25

  1+2sinxsiny+2cosxcosy+1=1

  2(sinxsiny+cosxcosy)=-1

  cos(x-y)=-1/2

  2.[sin50+cos40*(1+tan60*tan10)]/(cos20)^2

  ={sin50+cos40*[1+(sin60sin10/cos60cos10)]}/(cos20)^2

  ={sin50+cos40*[(cos60cos10/cos60cos10)+(sin60sin10/cos60cos10)]}/(cos20)^2

  ={sin50+cos40*[(cos60cos10+sin60sin10)/(cos60cos10)]}/(cos20)^2

  ={sin50+cos40*[cos(60-10)/(cos60cos10)]}/(cos20)^2

  =[sin50+(cos40cos50)/(cos60cos10)]/(cos20)^2

  ={sin50+[cos40cos(90-40)/(1/2)cos10]}/(cos20)^2

  =[sin(90-40)/(cos20)^2]+[2cos40sin40/cos10*(cos20)^2]

  =[(cos40)/(1+cos40)/2]+[(sin80)/cos10*(cos20)^2]

  =[2cos40/(1+cos40)]+[sin(90-10)/cos10*(cos20)^2]

  =[2cos40/(1+cos40)]+{(cos10)/cos10*[(1+cos40)/2]}

  =(2cos40+2)/(1+cos40)

  =2

  3.(2cos^2a-1)/[2tan*(45-a)*sin^2(45+a)]

  =(2cos^2a-1)/{2tan*(45-a)*[sin^2(90-(45-a)]}

  =(2cos^2a-1)/[2tan*(45-a)*cos^2(45-a)]

  =(cos2a)/[2sin(45-a)cos(45-a)]

  =(cos2a)/[sin(90-2a)]

  =(cos2a)/(cos2a)=1

  4.cos(a+b)=cosacosb-sinasinb=1/3

  cos(a-b)=cosacosb+sinasinb=1/6

  两式相加:2cosacosb=1/2

  两式相减:2sinasinb=-1/6

  tana*tanb=(sinasinb)/(cosacosb)=-1/3

  5.在三角形中:tanC=tan[π-(A+B)]=-tan(A+B)

  根据正切的两角和公式,变形后得:tanA+tanB={[tan(A+B)]*(1-tanAtanB)}

  展开:tanA+tanB=tan(A+B)-tan(A+B)tanAtanB

  将tanC=-tan(A+B)代入上式:tanA+tanB=-tanC+tanCtanAtanB

  整理后:tanA+tanB+tanC=tanAtanBtanC

  在三角形中:A+B+C=π,则:A/2+B/2+C/2=π/2

  tan(A/2)*tan(B/2)+tan(B/2)*tan(C/2)+tan(C/2)*tan(A/2)

  =tan(A/2)[tan(B/2)+tan(C/2)]+tan(B/2)*tan(C/2)

  ={tan[π/2-(B/2+C/2)]}*{[tan(B/2+C/2)][1-tan(B/2)*tan(C/2)]}+tan(B/2)*tan(C/2)

  =cot(B/2+C/2)*[tan(B/2+C/2)][1-tan(B/2)*tan(C/2)]+tan(B/2)*tan(C/2)

  =1[1-tan(B/2)*tan(C/2)]+tan(B/2)*tan(C/2)

  =1-tan(B/2)tan(C/2)+tan(B/2)*tan(C/2)

  =1

2020-06-24 15:29:26

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