How will you prove the formula #sin(A+B)=sinAcosB+cosAsinB# using formula of scalar product of two vectors?

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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#
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Sin(A+B) =SinA CosB + CosASinB formula can also be obtained
by taking scalar product of #hata and hat b#

Now

# hata* hatb=(cosAhati+sinAhatj)*(sinBhati+cosBhatj)#
#=>|hata||hatb|costheta=sinAcosB(hatj*hatj)+cosAsinB(hati*hati)#

Applying Properties of unit vectos #hati,hatj,hatk#
#hati*hatj=0 #
#hatj*hati=0 #
#hati*hati= 1 #
#hatj*hatj= 1#

and

#|hata|=1 and|hatb|=1#
Also inserting
#theta=90-(A+B)#,

Finally we get
#=>cos(90-(A+B))=sinAcosB+cosAsinB#

#:.sin(A+B)=sinAcosB+cosAsinB#