CVT1

Verify the identity: $\sin(a \pm b) = \sin(a)\cos(b)\pm\cos(a)\sin(b)\,$

Since $\sin(a) = \frac{e^{ai} - e^{-ai}}{2i}, \cos(b)=\frac{e^{bi} + e^{-bi}}{2}\,$

These two products can be computed:

$\sin(a)\cos(b) = \frac{e^{i(a+b)} + e^{i(a-b)} - e^{-i(a-b)} - e^{-i(a+b)}}{4i}\,$

$\cos(a)\sin(b) = \frac{e^{i(a+b)} - e^{i(a-b)} + e^{-i(a-b)} - e^{-i(a+b)}}{4i}\,$

$\sin(a)\cos(b) + \cos(a)\sin(b) = \frac{2e^{i(a+b)} - 2e^{-i(a+b)}}{4i} = \sin(a+b)\,$

$\sin(a)\cos(b) - \cos(a)\sin(b) = \frac{2e^{i(a-b)} - 2e^{-i(a-b)}}{4i} = \sin(a-b)\,$