已知tana=2 则tan(a-4 π)-搜答案-大师作文网

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

tanA-cosA=1（tanA-cosA）^2=1tan^2A+cot^2A-2=1tan^2A+cot^2A=3

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^2a+cot^2a=(tana+cota)²-2tana*cota=3²-2=7

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

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

3sinb=sin(2a+b)3sin(a+b-a)=sin(a+b+a)3[sin(a+b)cosa-cos(a+b)sina]=sin(a+b)cosa+cos(a+b)sinasin(a+b)c

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(

∵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

tan（a+b）=(tana+tanb)/(1-tanatanb)4=2/(1-tanatanb)所以tanatanb=1/2

tana=2tana/2/(1-[tana/2]^2)=4/(1-4)=-4/3tan(a+45)=(tana+tan45)/(1-tanatan45)=(tana+1)/(1-tana)=-1/3/

tan(a+β)=7(tana+tanβ)/(1-tanatanβ)=73(tana+tanβ)=73sin(a+β)/cosacosβ=7cosacosβ=3sin(a+β)/7=3(7/5√2)/