已知tan(a+b)=5分之3,tan(b-4分之π)=4分之1
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![已知tan(a+b)=5分之3,tan(b-4分之π)=4分之1](/uploads/image/f/4225316-68-6.jpg?t=%E5%B7%B2%E7%9F%A5tan%28a%2Bb%29%3D5%E5%88%86%E4%B9%8B3%2Ctan%28b-4%E5%88%86%E4%B9%8B%CF%80%29%3D4%E5%88%86%E4%B9%8B1)
因为cotB=1/3所以tanB=3tan(A+B)=(tanA+tanB)/(1-tanAtanB)=(4+3)/(1-4*3)=-7/11
你好~第一这道题考的是公式的变换,拆角呵呵.我差点就做不了,大二了,都混的差不多咯~回想高中真不错tan2A=tan=/=1/7tan2B的做法一样的,我就不做了第二:cosBAC=根号6/6,你就可
要证3sinB=sin(2A+B)即证3sin(A+B-A)=sin(A+B+A)即证3sin(A+B)cosA-3cos(A+B)sinA=sin(A+B)cosA+cos(A+B)sinA即证2s
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
5sinB=sin(2A+B)=sin(A+B+A)=sin(A+B)cosA+cos(A+B)sinA,sinB=sin(A+B-A)=sin(A+B)cosA-cos(A+B)sinA5sin(A
∵tan(b-π/3)=1/3∴tan(π/3-b)=-1/3∵tan(a+b)=3/5∴tan(a+π/3)=tan[(a+b)+(π/3-b)]=[tan(a+b)+tan(π/3-b)]/[1-
⑴tan[(A+B)/2]=tan[(A-B/2)+(B-A/2)]=(tan(A-B/2)+tan(B-A/2))/(1-tan(A-B/2)·tan(B-A/2))=(2-3)/1-(2·-3)=
1、tan2a=[tan(a+b)+tan(a-b)]/[1-tan(a+b)·tan(a-b)]=(3+2)/(1-3*2)=-1∴sin4a=2tan2a/[1+(tan2a)^2]=2(-1)/
因为cosA=4/5,且A是锐角,所以sinA=根号下[1-(cosA的平方)]=3/5,所以tanA=sinA/cosaA=3/4tanA=tan[(A-B)+B]=[tan(A-B)+tanB]/
tan(π-b)=1/3=-tanbtanb=-1/3tan2b=2tanb/(1-tanb^2)=-3/4sin(π-a)=3/5=sina,tana=-3/4tan(a+2b)=(tana+tan
(tanA+tanB)/(1-tanAtanB)=3(tanA-tanB)/(1+tanAtanB)==>2tanA-4tanAtanAtanB-4tanB+2tanAtanBtanB=0==>(si
sina=1/3cosa=1-(sina)2=1-1/9=8/9cos2a=2sinacosa=2*1/3*8/9=16/27tan(a-b)=(tana-tanb)/1-tanatanb=1/3ta
左边=(tana-tanb)/(-1/tana+cotb)=(tana-tanb)/(-1/tana+1/tanb)上下乘tanatanb=tanatanb(tana-tanb)/(tana-tanb
sin(a+b)=sinacosb+cosasinb=1/2sin(a-b)=sinacosb-cosasinb=1/3所以sinacosb=(1/2+1/3)/2=5/12cosasinb=(1/2
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+pai/4)=x用未知数x表示,比较简便tan(a+b)=tan[(a+pi/4)+(b-pi/4)]=[tan(a+pi/4)+tan(b-pi/a)]/[1-tan(a+pi/4)*t
tan2a=tan[(a+b)+(a-b)]=[tan(a+b)+tan(a-b)]/[1-tan(a+b)*tan(a-b)]=(3+5)/(1-3*5)=-4/7tan2b=tan[(a+b)-(
tan(A+B)=(tanA+tanB)/(1-tanAtanB)tan2a=tan[(a+b)+(a-b)]=(3+5)/(1-3*5)=-4/7tan(A-B)=(tanA-tanB)/(1+ta
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