已知tana=3,求tan(a π 4)的值

2a+b=(a+b)+ab=(a+b)-a为了简单设x=a+b3cos(2a+b)+5cosb=3cosxcosa-3sinxsina+5cosxcosa+5sinxsina=8cosxcosa+2s

2A+B=(A+B)+A,B=(A+B)-A,所以由SIN(2A+B)=3SINB得sin(A+B)cosA＋cos(A+B)sinA＝3sin(A+B)cosA-3cos(A+B)sinA2sin(

tan(a-π/4)=(tana-tanπ/4)/（1+tana*tanπ/4）=（2-1）/（1+2*1）=2/3

sin(2A+B)=sin(2A)cosB+cos(2A)sinB,sin(2A)=2tanA/(1+tanA^2)=1,cos(2A)=(1-tanA^2)/(1+tanA^2)=0,则由sin(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: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

(tanA-根3)（tanA-1）=0tanA=根3或tanA=1A=60度或A=45度

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

tana+1/tana=3可化成sina/cosa+cosa/sina=3化简得sin^2a+cos^2a/sinacosa=3可得出sinacosa=1/3由此可得出sina+cosa=根号15/3

3*sin(a+b-a)=sin(a+b+a);3*sin(a+b)cos(a)-3*cos(a+b)sin(a)=sin(a+b)cos(a)+cos(a+b)sin(a);2*sin(a+b)co

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,

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)=

tana=1a=kπ+π/43sinB=sin(2a+B)=sin(2kπ+π/2+B)=cosB3sinB=cosBsin^2B+cos^2B=1cos^2B=9/10sin^2B=1/10sinB

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+2b)=(tana+tan2b)/(1-tanatan2b)tan2b=2tanb/(1-tan^2b)tana=1/7,tanb=1/3带入tan2b=2tanb/(1-tan^2b)=

∵tanA-1/tanA=2∴平方,tan²A-2+1/tan²A=4∴tan²A+1/tan²A=6

2a-b=(a-b)+ab=(a-b)-a为了简单设x=a-b3cos(2a-b)+5cosb=3cosxcosa-3sinxsina+5cosxcosa+5sinxsina=8cosxcosa+2s