Following the last post, we now introduce the interesting consequences for the formula we derived. Remember we have these
sin(α+β)=sinαcosβ+sinβcosαsin(α−β)=sinαcosβ−sinβcosαcos(α+β)=cosαcosβ−sinβsinαcos(α−β)=cosαcosβ+sinβsinα
There are a few things we could do, for example, we could substitute β=α and get
sin(2α)=sin(α+α)=sinαcosα+sinαcosα=2sinαcosα
cos(2α)=cos(α+α)=cosαcosα−sinαsinα=cos2α−sin2α
More interestingly, we can add and subtract these formula together to do something, for example
sin(α+β)+sin(α−β)=sinαcosβ+sinβcosα+sinαcosβ−sinβcosα=sinαcosβ+sinαcosβ=2sinαcosβ
sin(α+β)−sin(α−β)=sinαcosβ+sinβcosα−sinαcosβ+sinβcosα=sinβcosα+sinβcosα=2sinβcosα
cos(α+β)+cos(α−β)=cosαcosβ−sinβsinα+cosαcosβ+sinβsinα=cosαcosβ+cosαcosβ=2cosαcosβ
cos(α−β)−cos(α+β)=cosαcosβ+sinβsinα−cosαcosβ+sinβsinα=sinβsinα+sinβsinα=2sinβsinα
If you look at these equation closely, you notice we converted a sum to a product, this can be very useful. To summarize, in this post, we have derived
sin(2α)=2sinαcosβcos(2α)=cos2α−sin2α
sin(α+β)+sin(α−β)=2sinαcosβsin(α+β)−sin(α−β)=2sinβcosαcos(α+β)+cos(α−β)=2cosαcosβcos(α−β)−cos(α+β)=2sinβsinα
sin(α+β)=sinαcosβ+sinβcosαsin(α−β)=sinαcosβ−sinβcosαcos(α+β)=cosαcosβ−sinβsinαcos(α−β)=cosαcosβ+sinβsinα
There are a few things we could do, for example, we could substitute β=α and get
sin(2α)=sin(α+α)=sinαcosα+sinαcosα=2sinαcosα
cos(2α)=cos(α+α)=cosαcosα−sinαsinα=cos2α−sin2α
More interestingly, we can add and subtract these formula together to do something, for example
sin(α+β)+sin(α−β)=sinαcosβ+sinβcosα+sinαcosβ−sinβcosα=sinαcosβ+sinαcosβ=2sinαcosβ
sin(α+β)−sin(α−β)=sinαcosβ+sinβcosα−sinαcosβ+sinβcosα=sinβcosα+sinβcosα=2sinβcosα
cos(α+β)+cos(α−β)=cosαcosβ−sinβsinα+cosαcosβ+sinβsinα=cosαcosβ+cosαcosβ=2cosαcosβ
cos(α−β)−cos(α+β)=cosαcosβ+sinβsinα−cosαcosβ+sinβsinα=sinβsinα+sinβsinα=2sinβsinα
If you look at these equation closely, you notice we converted a sum to a product, this can be very useful. To summarize, in this post, we have derived
sin(2α)=2sinαcosβcos(2α)=cos2α−sin2α
sin(α+β)+sin(α−β)=2sinαcosβsin(α+β)−sin(α−β)=2sinβcosαcos(α+β)+cos(α−β)=2cosαcosβcos(α−β)−cos(α+β)=2sinβsinα
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