18090所以sina/2>0,cosa/2<0sina=2sina/2cosa/2=-4/5sina/2=(-2/5)/cosa/2代入恒等式sin²(a/2)+cos²(a/2)=1所以解得cos²(a/2)=1/5或4/5所以cos(a/2)=-√5/5，sin(a/2)=2√5/5或cos(a/2)=-2√5/5，sin(a/2)=√5/5180同理得所以sin(b/2)>0,cosb<0cosb=2cos²(b/2)-1=3/5所以cos(b/2)=-2√5/5sin(b/2)=√5/5所以原式=sin(a/2)cos(b/2)-cos(a/2)sin(b/2)=-3/5或0
