若cos(a-b)cosa+sin(a-b)sina=-4/5,且b是第三象限角,则cos(b/2)的值为

cos(a-b)cosa+sin(a-b)sina
=cos[(a-b)-a]
=cos(-b)
=cosb=-4/5

b是第三象限角
2kπ+π<b<2kπ+3π/2
所以kπ+π/2<b/2<kπ+3π/4
若k是奇数，则k=2n+1,所以两边是2nπ+3π/2和2nπ+7π/4
b/2在第四象限
若k是偶数，则b/2在第二象限
所以 cosb/2正负都有可能

cosb=cos(b/2+b/2)=cosb/2*cosb/2-sinb/2*sinb/2=(cosb/2)^2-[1-(cosb/2)^2]
=2(cosb/2)^2-1
所以(cosb/2)^2=(-4/5+1)/2=1/10
cosb/2=±√10/10