已知0<a<π/4,0<b<π/4,3sinb=sin(2a+b),4tana/2=1

显示全部楼层 · 2011-10-10 00:11:58

解：因为4tan(a/2)=1-tan2(a/2)所以tana＝2tan(a/2)/[1-tan2(a/2)]=1/2又0<a<π/4，易解得sina＝√5/5,cosa=2√5/5则sin2a=2sinacosa=4/5，cos2a=1-2sin2a=3/5因为3sinb=sin(2a+b)所以3sinb=sin2acosb+cos2asinb即3sinb＝(4/5)cosb+(3/5)sinb(12/5)sinb=(4/5)cosb解得tanb=sinb/cosb=1/3所以tan(a+b)＝（tana+tanb)/(1-tana*tanb)
=(1/2 +1/...

千问 · 2011-10-10 00:11:58

解：4tan(a/2)=1-tan2(a/2)
2tan(a/2)/[1-tan2(a/2)]=1/2
tana=1/2
3sinb=sin(2a+b)
3sin[(a+b)-a]=sin[(a+b)+a]
3sin(a+b)cosa-3cos(a+b)sina=sin...