2a=b c a²tanB=sinBsinC 判断三角形形状

sin2a=2tana/[1+(tana)^2]=2*1/[1+1]=1(cos2a)^2=1-(sin2a)^2(cos2a)^2=1-1(cos2a)^2=0cos2a=03sinB=sin(2a

由tana=0.5得sin2a=2*0.5/(0.5^2+1)=4/5cos2a=(1-0.5^2)/(1+0.5^2)=3/5sin(2a+b)=sin2a*cosb+cos2a*sinb=4/5c

证明：⑴sin(A＋B)＝sinAcosB＋sinBcosA＝3/5sin(A－B)＝sinAcosB－sinBcosA＝1/5两式相加,得：2sinAcosB＝4/5sinAcosB＝2/5①则si

sin(A+B)=sinAcosB+cosAsinB=3/5sin(A-B)=sinAcosB-cosAsinB=1/5两式分别相加减,得sinAcosB=2/5cosAsinB=1/5两式相除tan

sinacosb+cosasinb=1/2sinacosb-cosasinb=1/3所以sinacosb=5/12cosasinb=1/12相除(sina/cosa)*(cosb/sinb)=5(si

tan(A-B)=(tanA-tanB)/(1+tanA*tanB)tan(A-B)/tanA+sin²C/sin²A=1左右移项得1-[(tanA-tanB)/(1+tanA*t

sin(a+b)=1cos(a+b)=0a+b=2kπ+π/2tan(2a+b)+tanb=tan[(a+b)+b]+tanb=-cotb+tan

sin(a+b)=1,cos(a+b)=0tan(2a+b)+tanb=tan[a+(a+b)]+tanb=[sinacos(a+b)+cosasin(a+b)]/[cosacos(a+b)-sina

把a看成（a+b-b)把（a+2b)看成（a+b+b)原式变化得sin(a+b)cosb+cos(a+b)sinb=3sin(a+b)cosb-3cos(a+b)sinb而tan(a+b)/tanb=

sin(a+2b)=3sinasin(a+b+b)=3sin(a+b-b)sin(a+b)cosb+cos(a+b)sinb=3sin(a+b)cosb-3cos(a+b)sinb-2sin(a+b)

（tana-tanb）/（tana+tanb）=（sina/cosa-sinb/cosb）/（sina/cosa+sinb/cosb）=(sinacosb-cosasinb)/(sinacosb+co

由sin(a+b)=sinacosb+cosasinbsin(a+b)=sinacosb+cosasinb可知sinacosb=[sin(a+b)+sin(a-b)]/2=5/24cosasinb=[

1sin(a+b)=2/3sina*cosb+cosa*sinb=2/3-----------(1)sin(a-b)=1/5sina*cosb-cosa*sinb=1/5-----------(2)联

如果题目没错的话.这比值是不定的.2sin(a+b-a)=sin(a+b-a)==>sin（a+b）cosa=3cos(a+b)sina==>tan(a+b)=3tanatanb=tan(a+b-a)

2sinb=sin(2a+b)令sinb=1/2sin(2a+b)=1b=30,2a+b=90a=30tan(a+b)/tanb=tan60/tan30=3

证明：sin(a+b)=1→cos(a+b)=√[1-sin^2(a+b)]=0→sin(2a+2b)=2*sin(a+b)*cos(a+b)=0→tan(2a+2b)=sin(2a+2b)/cos(

sin(A+2B)=sin[(A+B)+B}=sin(A+B)cosB+cos(A+B)sinBsinA=sin[(A+B)-B]=sin(A+B)cosB-cos(A+B)sinB代入原式中,得5s

把tan化成sin/cos,在化简得sinBcosB=sinAcosA,所以sin2B=sin2A,所以2B=2A或2B=180-2A所以A=B(等腰三角形)或A+B=90（直角三角形）

5sina=3sin(a-2B),5sin[(a-b)+b]=3sin[(a-b)-b]5sin(a-b)cosb+5cos(a-b)sinb=3sin(a-b)cosb-3cos(a-b)sinb2

sin(A+B)=1A+B=2kπ+π/22A+2B=4kπ+πtan(2A+2B)=tan(4kπ+π)=0tan[(2A+B)+B]=0所以[tan(2A+B)+tanB]/[1-tan(2A+B