Question

设向量a=(4cosa,sina),b=(sinB,4cosB),c=(cosB,-4sinB) 1)若a向量与b向量-

Answer 1

a向量与b向量-2c向量垂直 =>

a.(b-2c) =0

(4cosA,sinA).[(sinB,4cosB)+2(cosB,-4sinB)] =0

4cosA(sinB+2cosB) + sinA(4cosB-8sinB) =0

8(cosAcosB-sinAsinB)+ 4(sinAcosB+cosAsinB) =0

8cos(A+B)+4sin(A+B) =0

8+4tan(A+B) =0

tan(A+B) = -1/2