Question

已知向量a=(cosA,-2),b=(sinA,1)且向量a.b平行，则tan(A-圆周率/4)=??(急求帮忙，谢谢)

Answer 1

向量a=(cosA,-2),b=(sinA,1)且向量a.b平行，
所以
sinA/cosA=-1/2=tanA
tan(A-圆周率/4)=(tanA-tanπ/4)/(1+tanAtanπ/4)
=(-1/2-1)/(1-1/2)
=-3

Answer 2

向量a.b平行,则cosA/-2=sinA/1得：tanA= -1/2
tan(A-π/4)=(tanA-tanπ/4)/(1+tanA*tanπ/4)=(-3/2）/(1/2)= -3

Answer 3

因a//b，则：cosA/sinA=－1/2，即：tanA=－1/2，tan(A－π/4)=[tanA－tan(π/4)]/[1＋tanAtan(π/4)]=[tanA－1]/[1＋tanA]=－3

Answer 4

解：向量平行，对应坐标成比例，因为有数字，排除0向量可能。
cosA*1=sinA*-2
tanA=-1/2
tan(A-π/4)=(-1/2-1)/1+(-1/2)*1=-3

Answer 5

cosa/-2=sina/1
-0.5=sina/cosa
tana=-0.5
tan(a-π/4)=(tana-tanπ/4)/(1+tana*tanπ/4)=-1.5/0.5=-3