Question

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

Answer 1

As below

Explanation:

`
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`