已知cos(a+b)=-1/3,cos 2a=-5/13,a,b均为钝角,求sin(a-b)
问题描述：
已知cos(a+b)=-1/3,cos 2a=-5/13,a,b均为钝角,求sin(a-b)
答
cos(a+b)=-1/3,sin(a+b)=2根号2/3
cos2a=-5/13，sin2a=12/13
cos(a-b)=cos[2a-(a+b)]
=cos2acos(a+b)+sin2asin(a+b)
=5/39+24根号2/39
=（5+24根号2）/39
答
cos2a=-5/13
90