高中三角涵数,不是很难,已知sin(2a+b)=3sinb,设tana=x,tanb=y,记y=f(x).(1)求证：tan(a+b)=2tana(2)求f(x)的解析式

Question

高中三角涵数,不是很难,
已知sin(2a+b)=3sinb,设tana=x,tanb=y,记y=f(x).
(1)求证：tan(a+b)=2tana
(2)求f(x)的解析式

Answer 1

第一问：3sinB=sin（2A+B）
3sinB=sinAcos(A+B)+cosAsin(A+B)
3sinB=sinA(cosAcosB-sinAsinB)+cosA(sinAcosB+cosAsinB)
3sinB=sinAcosAcosB-sinAsinAsinB+cosAsinAcosB+cosAcosAsinB
3sinB=2cosAsinAcosB+sinB(cosAcosA-sinAsinA)
3(sinAsinA+cosAcosA)sinB=2cosAsinAcosB+sinB(cosAcosA-sinAsinA)
4sinAsinAsinB+2cosAcosAsinB=2cosAsinAcosB
2sinAsinAsinB+cosAcosAsinB=cosAsinAcosB
2sinAsinAsinB+cosAcosAsinB=2cosAsinAcosB-cosAsinAcosB
2sinAcosAcosB-2sinAsinAsinB=cosAsinAcosB+cosAcosAsinB
2sinA(cosAcosB-sinAsinB)=cosA(sinAcosB+cosAsinB)
2sinAcos(A+B)=cosAsin(A+B)
tan（A+B）=2tan A
第二问：sin(2a+b)=3sinb
sin[(a+b)+a]=3sin[(a+b)-a]
sin(a+b)cosa+cos(a+b)sina=3[sin(a+b)cosa-cos(a+b)sina]
sin(a+b)+cos(a+b)tga=3[sin(a+b)-cos(a+b)tga]
sin(a+b)+cos(a+b)=3[sin(a+b)-cos(a+b)]
4cos(a+b)=2sin(a+b)
tg(a+b)=2
tg(a+b)=(tga+tgb)/(1-tgatgb)=2
tgb+2tgatgb=2-tga
tgb=(2-tga)/(1+2tga)
y=(2-x)/(1+2x)