Question

Prove sin(a+b)=sinacosb+cosasinb by vector method

Answer 1

Let α⃗  and β⃗  be two unit vectors, and A and B be the angles made by them respectively with the X-axis.

So, α⃗ =cosAiˆ+sinAjˆβ⃗ =cosBiˆ+sinBjˆ

Now, α⃗ .β⃗ =(cosAiˆ+sinAjˆ)(cosBiˆ+sinBjˆ)

⇒αβcos(A−B)=cosAcosB+sinAsinB⇒cos(A−B)=cosAcosB+sinAsinB    [∵α=1, β=1] ------------(1)

Putting -B in place of B in (1):-

cos(A−(−B))=cosAcos(−B)+sinAsin(−B)⇒cos(A+B)=cosAcosB−sinAsinB


Similarly, 

α⃗ ×β⃗ =(cosAiˆ+sinAjˆ)×(cosBiˆ+sinBjˆ)⇒α⃗ ×β⃗ =cosAsinBkˆ−sinAcosBkˆ

⇒αβsin(A−B)(−kˆ)=(sinAcosB−cosAsinB)(−kˆ)⇒sin(A−B)=sinAcosB−cosAsinB −−−−−−−(2)

Putting B=-B in (2):-

sin(A−(−B))=sinAcos(−B)−cosAsin(−B)⇒sin(A+B)=sinAcosB+cosAsinB

Hence Proved.