Statement
When adding or subtracting angles in sine or cosine, these formulae apply for $ \alpha, \beta \in \R $:
$$ \sin(\alpha + \beta) = \sin(\alpha) \cos(\beta) + \cos(\alpha) \sin(\beta) $$
$$ \sin(\alpha - \beta) = \sin(\alpha) \cos(\beta) - \cos(\alpha) \sin(\beta) $$
$$ \cos(\alpha + \beta) = \cos(\alpha) \cos(\beta) - \sin(\alpha) \sin(\beta) $$
$$ \cos(\alpha - \beta) = \cos(\alpha) \cos(\beta) + \sin(\alpha) \sin(\beta) $$
Proof
Use Euler's formula:
$$ e^{\theta i} = \cos(\theta) + i\sin(\theta) $$
When substituting $ \alpha + \beta $ for $ \theta $, you get:
$$ e^{(\alpha + \beta)i} = \cos(\alpha + \beta) + i\sin(\alpha + \beta) $$
Now, take $ e^{(\alpha + \beta)i} $ again and expand the brackets. Use that $ i^2 = -1 $.
$$\begin{aligned}e^{(\alpha + \beta)i} &= e^{\alpha i + \beta i} \newline&= e^{\alpha i} \cdot e^{\beta i} \newline&= \bigg(\cos(\alpha) + i\sin(\alpha)\bigg) \cdot \bigg(\cos(\beta) + i\sin(\beta)\bigg) \newline&= \cos(\alpha) \cos(\beta) + i \cos(\alpha) \sin(\beta) + i \sin(\alpha) \cos(\beta) - \sin(\alpha) \sin(\beta) \newline&= \cos(\alpha) \cos(\beta) - \sin(\alpha) \sin(\beta) + i \bigg(\cos(\alpha) \sin(\beta) + \sin(\alpha) \cos(\beta)\bigg)\end{aligned}$$
Since both expressions are equal to $ e^{(\alpha + \beta)i} $, set them equal to eachother:
$$ \cos(\alpha + \beta) + i \sin(\alpha + \beta) = \cos(\alpha) \cos(\beta) - \sin(\alpha) \sin(\beta) + i \bigg(\cos(\alpha) \sin(\beta) + \sin(\alpha) \cos(\beta)\bigg) $$
Since sine and cosine give a real value for a real input, the real parts and imaginary parts must match, giving:
$$ \cos(\alpha + \beta) = \cos(\alpha) \cos(\beta) - \sin(\alpha) \sin(\beta) $$
$$ \sin(\alpha + \beta) = \cos(\alpha) \sin(\beta) + \sin(\alpha) \cos(\beta) $$
When substituting $ - \beta $ for $ \beta $, use that $ \sin(-\theta) = -\sin(\theta) $ and $ \cos(-\theta) = \cos(\theta) $:
$$ \cos(\alpha - \beta) = \cos(\alpha) \cos(- \beta) - \sin(\alpha) \sin(-\beta) = \cos(\alpha) \cos(\beta) + \sin(\alpha) \sin(\beta) $$
$$ \sin(\alpha - \beta) = \cos(\alpha) \sin(- \beta) + \sin(\alpha) \cos(- \beta) = \sin(\alpha) \cos(\beta) - \cos(\alpha) \sin(\beta) $$