已知向量a=(cosa,sina),b=(根号2-sina,cosa),a∈(-π/2,π/2)

（a-c)*b=(cosa-√2)*(√2-sina)=√2(sina+cosa)-sinacosa-2设(sina+cosa)=t,则可得：sinacosa=(t^2-1)/2,且t∈(-1,√2]所以,得：（a-c)*b=-t^2/2+√2t-3/2=-(t-√2)^2/2-1/2所以,当t=√2时,（a-c)*b的最大值=-1/2....好球