已知tana=1 2,tan(a-b)求
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tanA=tan[(B+A)-B]=[tan(B+A)-tanB]/[1+tan2BtanB]=tanB/[1+2(tanB)^2]=1/[1/tanB+2tanB]≤1/(2√2)=√2/4
tan(a-π/4)=(tana-tanπ/4)/(1+tana*tanπ/4)=(2-1)/(1+2*1)=2/3
tan(π/4+A)=sin(π/4+A)/cos(π/4+A)=(sinπ/4*cosA+cosπ/4*sinA)/(cosπ/4*cosA-sinπ/4*sinA)=(tanπ/4*cosA+si
将tan(a+b)化简,易知tana*tanb=1/2
Tan(a)=Tan(a/2+a/2)=(2Tana/2)/(1-(Tana/2)^2)=-4/3
tan(a+π/4)=(tana+tanπ/4)/1-tana*tanπ/4=(2+1)/1-2*1=-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
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
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
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
tan(a+β)=7(tana+tanβ)/(1-tanatanβ)=73(tana+tanβ)=73sin(a+β)/cosacosβ=7cosacosβ=3sin(a+β)/7=3(7/5√2)/
1.∵tan(a/2)=2∴tana=[2tan(a/2)]/{1-[tan(a/2)]^2}=(2×2)/(1-2^2)=-4/3∴tan(a+π/4)=[tana+tan(π/4)]/[1-tan
结果:tana*tanb=1/2.过程也不复杂,把tana移项,然后展开tan(a+b),再全部通分,两边合并同类项.
tan(90度-a)=cota=1/tana=1/2再问:你和我的答案一样哦,只是我不确定,谢谢啊
分析法倒推tanr=-tan(a-r-a)=[tana-tan(a-r)]/[1+tana*tan(a-r)]tana*tanr=[tan^2a-tana*tan(a-r)]/[1+tana*tan(
tan3a=[3tana-(tana)^3]/[1-3(tana)^2]=1/tana,∴3(tana)^2-(tana)^4=1-3(tana)^2,∴(tana)^4-6(tana)^2+1=0,
∵tanA-1/tanA=2∴平方,tan²A-2+1/tan²A=4∴tan²A+1/tan²A=6
应该是tan(π/4-a)=(tanπ/4-tana)/(tanπ/4+tana)=(1-tana)/(1+tana)=根号5
tana=2tana/2/(1-[tana/2]^2)=4/(1-4)=-4/3tan(a+45)=(tana+tan45)/(1-tanatan45)=(tana+1)/(1-tana)=-1/3/