Rappel :

• $(a + b)^3 = a^3 + b^3 + 3ab(a + b)$
• $(a - b)^3 = a^3 - b^3 - 3ab(a - b)$

Calculons $(\cos x + i \sin x)^3 = \cos(3x) + i \sin(3x)$ :

$$\begin{aligned} (\cos x + i \sin x)^3 &= \cos^3 x + i^3 \sin^3 x + 3i \cos x \sin x (\cos x + i \sin x) \\ &= \cos^3 x - i \sin^3 x + 3i \cos^2 x \sin x - 3 \cos x \sin^2 x \\ &= \cos^3 x - 3 \cos x \sin^2 x + i (-\sin^3 x + 3 \cos^2 x \sin x) \end{aligned}$$

D'où :

$$\begin{cases} \cos(3x) = \cos^3 x - 3 \cos x \sin^2 x \\ \sin(3x) = -\sin^3 x + 3 \cos^2 x \sin x \end{cases}$$

Or, on sait que $\cos^2 x + \sin^2 x = 1$, donc :

$$\begin{cases} \cos^2 x = 1 - \sin^2 x \\ \sin^2 x = 1 - \cos^2 x \end{cases}$$

Ainsi, on obtient :

$$\begin{cases} \cos(3x) = 4 \cos^3 x - 3 \cos x \\ \sin(3x) = 3 \sin x - 4 \sin^3 x \end{cases}$$