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

1 Answer
P dilip_k
Jun 10, 2016

As below

Explanation:

self draw

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

  • hata->ˆa→ inclined with positive direction of X-axis at angles A
  • hat b->ˆb→ inclined with positive direction of X-axis at angles 90+B, where 90+B>A90+B>A
  • Angle between these two vectors becomes
    theta=90+B-A=90-(A-B)θ=90+B−A=90−(A−B),

hata=cosAhati+sinAhatjˆa=cosAˆi+sinAˆj
hatb=cos(90+B)hati+sin(90+B)ˆb=cos(90+B)ˆi+sin(90+B)
=-sinBhati+cosBhatj=−sinBˆi+cosBˆj
Now
hata xx hatb=(cosAhati+sinAhatj)xx(-sinBhati+cosBhatj)ˆa׈b=(cosAˆi+sinAˆj)×(−sinBˆi+cosBˆj)
=>|hata||hatb|sinthetahatk=cosAcosB(hatixxhatj)-sinAsinB(hatjxxhati)⇒|ˆa|∣∣ˆb∣∣sinθˆk=cosAcosB(ˆi׈j)−sinAsinB(ˆj׈i)
Applying Properties of unit vectos hati,hatj,hatkˆi,ˆj,ˆk
hatixxhatj=hatk ˆi׈j=ˆk
hatjxxhati=-hatk ˆj׈i=−ˆk
hatixxhati= "null vector" ˆi׈i=null vector
hatjxxhatj= "null vector" ˆj׈j=null vector
and
|hata|=1 and|hatb|=1" ""As both are unit vector" |ˆa|=1and∣∣ˆb∣∣=1 As both are unit vector

Also inserting
theta=90-(A-B)θ=90−(A−B),

Finally we get
=>sin(90-(A-B))hatk=cosAcosBhatk+sinAsinBhatk⇒sin(90−(A−B))ˆk=cosAcosBˆk+sinAsinBˆk

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

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