Prove, by vector method, that sin (α + β) = sin α . cos β + cos α . sin β

#### Solution

Let ∠XOP and ∠XOQ be in standard position and m∠XOP = - α ,m∠XOQ = β

Take a point A on ray OP and a point B on ray OQ such that OA = OB = 1.

Since cos (- α) = cos α

and sin (- α) = - sin α,

A is (cos (- α), sin (- α)),

i.e. (cos α, - sin α)

B is (cos β, sin β)

∴ `bar"OA" = ("cos" alpha)bar"i" - ("sin" alpha).bar"j" + 0.bar"k"`

`bar"OB" = ("cos" beta)bar"i" - ("sin" beta).bar"j" + 0.bar"k"`

`∴ bar"OA" xx bar"OB" = |(hat"i",hat"j",hat"k"),("cos" alpha, - "sin" alpha, 0),("cos" beta, "sin" beta, 0)|`

= (cos α sin β + sin α cos β)`bar"k"` ....(1)

The angle between `bar"OA" "and" bar"OB"` is α + β.

Also, `bar"OA", `bar"OB"` lie in the XY-plane.

∴ the unit vector perpendicular to `bar"OA"` and `bar"OB"` is `bar"k"`.

∴ `bar"OA" xx bar"OB" = ["OA"."OB" "sin"(alpha + beta)]bar"k"`

= sin (α + β) . `bar"k"` ...(2)

∴ from (1) and (2),

sin (α + β) = sin α cos β + cos α sin β