Some trigonometry formula (II)

Following the last post, we now introduce the interesting consequences for the formula we derived. Remember we have these


$ \begin{eqnarray*} \sin(\alpha + \beta) &=& \sin \alpha \cos \beta + \sin \beta \cos \alpha \\ \sin(\alpha - \beta) &=& \sin \alpha \cos \beta - \sin \beta \cos \alpha \\ \cos(\alpha + \beta) &=& \cos \alpha \cos \beta - \sin \beta \sin \alpha \\ \cos(\alpha - \beta) &=& \cos \alpha \cos \beta + \sin \beta \sin \alpha \end{eqnarray*} $

There are a few things we could do, for example, we could substitute $ \beta = \alpha $ and get

$ \begin{eqnarray*} & & \sin(2\alpha) \\ &=& \sin(\alpha + \alpha) \\ &=& \sin \alpha \cos \alpha + \sin \alpha \cos \alpha \\ &=& 2 \sin \alpha \cos \alpha \\ \end{eqnarray*} $

$ \begin{eqnarray*} & & \cos(2\alpha) \\ &=& \cos(\alpha + \alpha) \\ &=& \cos \alpha \cos \alpha - \sin \alpha \sin \alpha \\ &=& \cos^2 \alpha - \sin^2 \alpha \end{eqnarray*} $

More interestingly, we can add and subtract these formula together to do something, for example

$ \begin{eqnarray*} & & \sin(\alpha + \beta) + \sin(\alpha - \beta) \\ &=& \sin \alpha \cos \beta + \sin \beta \cos \alpha + \sin \alpha \cos \beta - \sin \beta \cos \alpha \\ &=& \sin \alpha \cos \beta + \sin \alpha \cos \beta \\ &=& 2 \sin \alpha \cos \beta \\ \end{eqnarray*} $

$ \begin{eqnarray*} & & \sin(\alpha + \beta) - \sin(\alpha - \beta) \\ &=& \sin \alpha \cos \beta + \sin \beta \cos \alpha - \sin \alpha \cos \beta + \sin \beta \cos \alpha \\ &=& \sin \beta \cos \alpha + \sin \beta \cos \alpha \\ &=& 2 \sin \beta \cos \alpha \\ \end{eqnarray*} $

$ \begin{eqnarray*} & & \cos(\alpha + \beta) + \cos(\alpha - \beta) \\ &=& \cos \alpha \cos \beta - \sin \beta \sin \alpha + \cos \alpha \cos \beta + \sin \beta \sin \alpha \\ &=& \cos \alpha \cos \beta + \cos \alpha \cos \beta \\ &=& 2 \cos \alpha \cos \beta \\ \end{eqnarray*} $

$ \begin{eqnarray*} & & \cos(\alpha - \beta) - \cos(\alpha + \beta) \\ &=& \cos \alpha \cos \beta + \sin \beta \sin \alpha - \cos \alpha \cos \beta + \sin \beta \sin \alpha \\ &=& \sin \beta \sin \alpha + \sin \beta \sin \alpha \\ &=& 2 \sin \beta \sin \alpha \end{eqnarray*} $

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

$ \begin{eqnarray*} \sin(2\alpha) &=& 2 \sin \alpha \cos \beta \\ \cos(2\alpha) &=& \cos^2 \alpha - \sin^2 \alpha \\ \end{eqnarray*} $

$ \begin{eqnarray*} \sin(\alpha + \beta) + \sin(\alpha - \beta) &=& 2 \sin \alpha \cos \beta \\ \sin(\alpha + \beta) - \sin(\alpha - \beta) &=& 2 \sin \beta \cos \alpha \\ \cos(\alpha + \beta) + \cos(\alpha - \beta) &=& 2 \cos \alpha \cos \beta \\ \cos(\alpha - \beta) - \cos(\alpha + \beta) &=& 2 \sin \beta \sin \alpha \\ \end{eqnarray*} $