已知tan(3.14+a)=-1/3,若a是钝角,a-b是锐角,且sin(a-b)=3/5,求sinb

问题描述:

已知tan(3.14+a)=-1/3,若a是钝角,a-b是锐角,且sin(a-b)=3/5,求sinb

∵tan(π+a)=tana=-1/3,且a是钝角∴sina=√10/10 ,cosa= -3√10/10又∵a-b是锐角∴cos(a-b)=4/5∴sinb = sin[a - (a-b)] = sinacos(a-b) - cosasin(a-b)=(√10/10)×(4/5) - (-3√10/10)×(3/5)=13√10/50...