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

2 Answers
May 30, 2016

Answer:

I could prove it using the dot product of vectors.

Explanation:

Let
#hatA and hatB# be two unit vectors in the #x#-#y# plane such that #hatA# makes an angle #-A# and #hatB# makes an angle #B# with #x#-axis so that the angle between the two is #(A+B)#
The unit vectors can be written in Cartesian form as
#hatA =cosAhat i- sin A hat j# and #hatB =cosBhat i +sin B hat j# ....(1)
To prove
#cos(A+B)=cosAcosB−sinAsinB#

We know that dot product of two vectors is
#vecA cdot vecB=|vecA|| vecB|cos theta#
Inserting our unit vectors in the above; #|vecA|=| vecB|=1# and value of #theta=(A+B)#, we obtain

#hatA cdot hatB=cos (A+B)#
Using equation (1)
LHS #=(cosAhat i- sin A hat j)cdot (cosBhat i +sin B hat j)#
From property of dot product we know that only terms containing #haticdothati and hatjcdothatj " are" =1# and rest vanish.
#:.# LHS#=cosAcosB-sin Asin B#

Equating LHS with RHS we obtain

#cos(A+B)=cosAcosB−sinAsinB#

Jun 2, 2016

Answer:

As follows

Explanation:

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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#