Question #683a7

1 Answer
Mar 19, 2017

drawn

````````````````````````````````````````````````````````````````````````````````````````````````````````````````
Let us consider two unit vectors in X-Y plane as follows :

  • #hata-># inclined with positive direction of X-axis at angles A
  • # hat b-># inclined with positive direction of X-axis at angles (90-B), where # (90-B)>A#
  • Angle between these two vectors becomes
    #theta=90-B-A=90-(A+B)#,

#hata=cosAhati+sinAhatj#
#hatb=cos(90-B)hati+sin(90-B)#
#=sinBhati+cosBhatj#
Now
# hata xx hatb=(cosAhati+sinAhatj)xx(sinBhati+cosBhatj)#
#=>|hata||hatb|sinthetahatk=cosAcosB(hatixxhatj)+sinAsinB(hatjxxhati)#
Applying Properties of unit vectos #hati,hatj,hatk#
#hatixxhatj=hatk #
#hatjxxhati=-hatk #
#hatixxhati= "null vector" #
#hatjxxhatj= "null vector" #
and
#|hata|=1 and|hatb|=1" ""As both are unit vector" #

Also inserting
#theta=90-(A+B)#,

Finally we get
#=>sin(90-(A+B))hatk=cosAcosBhatk-sinAsinBhatk#

#:.cos(A+B)=cosAcosB-sinAsinB#