Trigonometria: transformação de soma em produto.

Sabemos que:

$\sin (a + b) = (\sin a)(\cos b) + (\sin b)(\cos a)$ (I)

$\sin (a - b) = (\sin a)(\cos b) - (\sin b)(\cos a)$ (II)

$\cos (a + b) = (\cos a)(\cos b) - (\sin a)(\sin b)$ (III)

$\cos (a - b) = (\cos a)(\cos b) + (\sin a)(\sin b)$ (IV)

Somando (I) e (II): $2(\sin a)(\cos b) = \sin (a + b) + \sin (a - b)$.

Subtraindo (II) de (I): $2(\sin b)(\cos a) = \sin (a + b) - \sin (a - b)$.

Somando (III) e (IV): $2(\cos a)(\cos b) = \cos (a + b) + \cos (a - b)$.

Subtraindo (IV) de (III): $-2(\sin a)(\sin b) = \cos (a + b) - \cos (a - b)$.

Fazendo $p = a + b$ e $q = a - b$, teremos que $a = \dfrac{p + q}{2}$ e $b = \dfrac{p - q}{2}$. Substituindo:

$\sin p + \sin q = 2\left(\sin \dfrac{p + q}{2}\right)\left(\sin \dfrac{p - q}{2}\right)$

$\sin p - \sin q = 2\left(\cos \dfrac{p + q}{2}\right)\left(\sin \dfrac{p - q}{2}\right)$

$\cos p + \cos q = 2\left(\cos \dfrac{p + q}{2}\right)\left(\cos \dfrac{p - q}{2}\right)$

$\cos p - \cos q = -2\left(\sin \dfrac{p + q}{2}\right)\left(\sin \dfrac{p - q}{2}\right)$