s8皮城符文:已知tga=1 sin(2a+b)=3sinb 求tg(a+b)的值

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∵tga=1,∴a=kπ+π/4,k∈z.∴2a=2kπ+π/2
∴sin(2a+b)=sin2acosb+sinbcos2a=cosb=3sinb
∴tanb=1/3
∴tg(a+b)=(tga+tgb)/(1-tgatgb)=(1+1/3)/(1-1/3)=2

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

sin(2a+b)=3sinb =>sin(2a+b)-sinb=2sinb
=>2cos〔(2a+b+b)/2〕*sin〔(2a+b-b)/2〕=2sinb 和差化积
=>cos(a+b)*sina=sinb=>cos(a+b)=sinb/sina
sin(2a+b)=3sinb =>sin(a+b)cosa+cos(a+b)sina=3sinb
=>sin(a+b)cosa+(sinb/sina)*sina=3sinb
=>sin(a+b)cosa=2sinb =>sin(a+b)=2sinb/cosa
tg(a+b)=sin(a+b)/cos(a+b)=(2sinb/cosa)/(sinb/sina)=(2sinb/cosa)*(sina/sinb)=2(sina/cosa)=2tga=2

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