已知cosa=-1/2,sin(a-b)=3/5,且a属于(90,180),a-b属于(0,90),求sinb

已知：cosa=-1/2,sin(a-b)=3/5,且a属于(90,180),a-b属于(0,90),那么：
sina=√(1-cos²a)=(√3)/2,cos(a-b)=√[1-sin²(a-b)]=4/5
sinb=sin[a-(a-b)]
=sina*cos(a-b)- cosa*sin(a-b)
=(√3)/2 *(4/5) - (-1/2)*(3/5)
=(4√3+3)/10