已知tan(a b)=2 tanb=三分之一
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将tan(a+b)化简,易知tana*tanb=1/2
因为tanb=1/3所以tan(2b)=2tanb/(1-tanbtanb)=3/4,所以tan(a+2b)=(tana+tan2b)/(1-tanatan2b),将tana=1/7和tan2b=3/
∵tanA=1,tanB=2,∴tan(A+B)=tanA+tanB1−tanAtanB=1+21−2=-3,故答案为:-3.
tan(a+b)=4(tana+tanb)/(1-tanatanb)=4tana+tanb=2(1)所以2/(1-tanatanb)=4所以tanatanb=1/2(2)由(1)(2)tana,tan
tan(A+B)=4(tanA+tanB)/(1-tanAtanB)=4tanAtanB=1/2再联立tanA+tanB=2又tanA
tanA:tanB:tanC=1:2:3→tanA:tanB:tan【π-(A+B)】=1:2:3→tanA:tanB:—tan(A+B)=1:2:3,3tanA=—tan(A+B)=—(tanA+t
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
cos(a+b)=cosacosb-sinasinb=0---->sinasinb=cosacosb两边同时除以cosacosb,得tanatanb=1所以tan(2a+b)+tanb=0
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
(1)cos2a+sinacosa+cos²a=0∴cos²a-sin²a+sinacosa+cos²a=0即2cos²a+sinacosa-sin&
tan(A+B)=tan(π-C)=tanC=1C=π/4,或C=3π/4tanA=tan[(A+B)-B]=[tan(A+B)-tanB]/[1+tan(A+B)*tanB]=1/2tanB
可能我算错了我算的是tan(A-B)=sin2B/(3-cos2B)下面是我算的tan(A-B)=(tanA-tanB)/(1+tanAtanB)=tanB/[1+(tanB)^2]=sinAcosA
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(
tan(a+2b)=(tana+tan2b)/(1-tanatan2b)tan2b=2tanb/(1-tan^2b)tana=1/7,tanb=1/3带入tan2b=2tanb/(1-tan^2b)=
tan(a+b)=(-7-3)/(21-1)=-1/2tan(2a+b)=[tana+tan(a+b)]/[1-tana*tan(a+b)]=(-2-3)/(6-1)=-1
tan(a+b)=(tana+tanb)/(1-tanatanb)=(2/5+3/7)/(1-2/5*3/7)=(14/35+15/35)/(1-6/35)=(29/35)/(29/35)=1
tan(a+b)=(tana+tanb)/(1-tanatanb)=(2/5+3/7)/(1-2/5*3/7)=(14/35+15/35)/(1-6/35)=(29/35)/(29/35)=1tan(
tan(a+b)=(tana+tanb)/(1-tanatanb)4=2/(1-tanatanb)所以tanatanb=1/2