作业帮SINA+SINB+SINC=COSA+COSB+COSC=0,求TAN(A+B+C)+TANA*TANB*TANC
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SINA+SINB+SINC=COSA+COSB+COSC=0,求TAN(A+B+C)+TANA*TANB*TANC
SINA+SINB+SINC=COSA+COSB+COSC=0,求TAN(A+B+C)+TANA*TANB*TANC
SINA+SINB+SINC=COSA+COSB+COSC=0,求TAN(A+B+C)+TANA*TANB*TANC
-sinc=sina+sinb
-cosc=cosa+cosb
1=sinc^2+cosc^2=1+1+2sinasinb+2cosacosb=2+2cos(a-b)
cos(a-b)=-1/2
同理：
cos(b-c)=cos(a-c)=-1/2
b=a+120°(+360°的整数倍),c=a-120°(+360°的整数倍)
tan(a+b+c)=tan3a
=tan(2a+a)=(tan2a+tana)/(1-tan2atana)
=(2tana+tana(1-tana^2))/(1-tana^2-2tana^2)
=(3tana-tana^3)/(1-3tana^2)
tanatanbtanc
=tanatan(a+120)tan(a-120)
=tana(tana^2-3)/(1-3tana^2)
=(tana^3-3tana)/(1-3tana^2)
所以：
tan(a+b+c)+tanatanbtanc=0.
-sinc=sina+sinb -cosc=cosa+cosb 1=sinc^2+cosc^2=1+1+2sinasinb+2cosacosb=2+2cos(a-b) cos(a-b)=-1/2 同理： cos(b-c)=cos(a-c)=-1/2 b=a+120°(+360°的整数倍), c=a-120°(+360°的整数倍) tan(a+b+c)=tan3a ...
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-sinc=sina+sinb -cosc=cosa+cosb 1=sinc^2+cosc^2=1+1+2sinasinb+2cosacosb=2+2cos(a-b) cos(a-b)=-1/2 同理： cos(b-c)=cos(a-c)=-1/2 b=a+120°(+360°的整数倍), c=a-120°(+360°的整数倍) tan(a+b+c)=tan3a =tan(2a+a)=(tan2a+tana)/(1-tan2atana) =(2tana+tana(1-tana^2))/(1-tana^2-2tana^2) =(3tana-tana^3)/(1-3tana^2) tanatanbtanc =tanatan(a+120)tan(a-120) =tana(tana^2-3)/(1-3tana^2) =(tana^3-3tana)/(1-3tana^2) 所以： tan(a+b+c)+tanatanbtanc=0.
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