10.2 CHAPTER 10. TRIGONOMETRY
Derivation ofsin(α−β) EMCCH
We can use
sin(α +β) = sinα cosβ + cosα sinβ
to show that
sin(α−β) = sinα cosβ− cosα sinβ
We know that
sin(−θ) =− sin(θ)
and
cos(−θ) = cosθ
Therefore,
sin(α−β) = sin(α + (−β))
= sinα cos(−β) + cosα sin(−β)
= sinα cosβ− cosα sinβDerivation ofcos(α+β) EMCCI
We can use
sin(α−β) = sinα cosβ− sinβ cosα
to show that
cos(α +β) = cosα cosβ− sinα sinβ
We know that
sin(θ) = cos(90−θ).
Therefore,
cos(α +β) = sin(90− (α +β))
= sin((90−α)−β))
= sin(90−α) cosβ− sinβ cos(90−α)
= cosα cosβ− sinβ sinαDerivation ofcos(α−β) EMCCJ
We found this identity inour derivation of the sin(α +β) identity. We can also use the fact that
sin(α +β) = sinα cosβ + cosα sinβto derive that
cos(α−β) = cosα cosβ + sinα sinβ
As
cos(θ) = sin(90−θ),