Formules de transformation produit - somme
Des formules d'addition nous tirons les produits :
\(\boxed{{\qquad\cos a \cos b = \frac{1}{2} [\cos(a+b) + \cos(a-b)]\quad\\ ~\qquad\sin a \sin b = -\frac{1}{2} [\cos(a+b) - \cos(a-b)]\\ ~\qquad \sin a \cos b = \frac{1}{2} [\sin(a+b) + \sin(a-b)] \\ ~\qquad\cos a \sin b = \frac{1}{2} [\sin(a+b) - \sin(a-b)]\quad}}\)
Démonstration :
Des formules d'addition :
\(\cos(a+b) = \cos a ~\cos b - \sin a ~\sin b \) (1)
\(\cos(a-b) = \cos a ~\cos b + \sin a ~\sin b \) (2)
\(\sin(a+b) = \sin a ~\cos b + \cos a ~\sin b \) (3)
\(\sin(a-b) = \sin a ~\cos b - \cos a ~\sin b \) (4)
On déduit par :
addition de (1) et (2) : \(\cos (a + b) + \cos (a - b) = 2 \cos a \cos b\)
\(\cos a \cos b = \frac{1}{2} [\cos(a+b) + \cos(a-b)]\)
soustraction de (1) et (2) : \(\cos (a + b) - \cos (a - b) = -2 \sin a \sin b\)
\(\sin a \sin b = -\frac{1}{2} [\cos(a+b) - \cos(a-b)]\)
addition de (3) et (4) : \(\sin (a + b) + \sin (a - b) = 2 \sin a \cos b\)
\(\sin a \cos b = \frac{1}{2} [\sin(a+b) + \sin(a-b)]\)
soustraction de (3) et (4) : \(\sin (a + b) - \sin (a - b) = 2 \cos a \sin b\)
\(\cos a \sin b = \frac{1}{2} [\sin(a+b) - \sin(a-b)]\)
Exemple :
Cas où \(a = x\) et \(b = 2x\)
\(\cos x \cos 2x = \frac{1}{2} [\cos 3x + \cos(-x)] = \frac{1}{2} (\cos 3x + \cos x)\)
\(\sin x \sin 2x = -\frac{1}{2} [\cos 3x - \cos(-x)] = -\frac{1}{2} (\cos 3x - \cos x)\)
\(\sin x \cos 2x = \frac{1}{2} [\sin 3x + \sin(-x)] = \frac{1}{2} (\sin 3x - \sin x)\)
\(\cos x \sin 2x = \frac{1}{2} [\sin 3x - \sin(-x)] = \frac{1}{2} (\sin 3x + \sin x)\)