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

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

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
`

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