How will you prove the formula #cos(A-B)=cosAcosB+sinAsinB# using formula of vector product of two vectors?

1 Answer
Jun 10, 2016

Answer:

As below

Explanation:

self draw

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#