若这两个向量共线则向量a-2b=0;
←→tan(a+b)-2tana=0; tan(a+b)=2tana
←→cosa·sin(a+b)=2sina·cos(a+b)
←→cosa·sin(a+b)+sina·cos(a+b)=3sina·cos(a+b)
←→sin[a+(a+b)]=3sina·cos(a+b); sin(2a+b)=3sina·cos(a+b);
←→3Sinb=3sina·cos(a+b);
sinb=sina·cos(a+b)=sina·(cosa·cosb-sina·sinb)=sina·cosa·cosb-sin^2a·sinb
=(1/2)sin2a·cosb-(1/2)(1-cos2a)·sinb
=(1/2)(sin2a·cosb+cos2a·sinb)-(1/2)·sinb
=(1/2)sin(2a+b) -(1/2)·sinb
=(1/2)Sinb-(1/2)·sinb
=0
所以反推成立;
那么正推一定也是成立的.
证明完毕