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