设已知向量a=(2cos(a+b)/2,sin(a-b)/2),b=(cos(a+b)/2,3sin(a-b)/2),a,b属于(0.π).若a*b=5/2,求tanatan
问题描述：
设已知向量a=(2cos(a+b)/2,sin(a-b)/2),b=(cos(a+b)/2,3sin(a-b)/2),a,b属于(0.π).若a*b=5/2,求tanatan
答
a*b=2[cos(a+b)/2]^2+3[sin(a-b)/2]^2=1+cos(a+b)+3/2-(3/2)sin(a-b)=5/2,所以cos(a+b)=(3/2)cos(a-b),即2cos(a+b)=3cos(a-b),即2cosacosb-2sinasinb=3cosacosb+3sinasinb,即5sinasinb=-cosacosb,所以tanatanb=-1/5....