Question

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

Answer 1

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`

Answer 2

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`