Some trigonometry formula (I)

In this post, we assume a few formula and derive some interesting trigonometry formula:

Assumed equations:

$\sin(\alpha + \beta) = \sin \alpha \cos \beta + \sin \beta \cos \alpha$
$\sin(-x) = -\sin(x)$
$\cos(-x) = \cos(x)$
$\cos(\alpha + \frac{\pi}{2}) = -\sin(\alpha)$
$\sin(\alpha + \frac{\pi}{2}) = \cos(\alpha)$

Let's get started:

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

So we obtain $\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$, next, try subtraction

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

And also, we have

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

So we obtain the subtraction formula as well, as a short summary, we have

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