Trigonométrie circulaire - Trigonométrie hyperbolique

1. Connaissant l'expression de \(\tan 2x = \frac{2 \tan x}{1-\tan^{2}x}\) (1pt), nous avons :

\(\begin{array}{ll}\tan 3x &= \tan(x+2x) = \frac{\tan x + \tan 2x}{1-(\tan x)(\tan 2x)} \\ &= \frac{\tan x (1-\tan^{2} x) + 2 \tan x}{(1-\tan^{2}x) - 2 \tan^{2}x} \end{array}\)

donc

\(\tan 3x = \frac{3 \tan x - \tan^{3}x}{1-3 \tan^{2}x}\) (2pts)

2. Sachant que \(\tan\frac{2 \pi}{3} = -\sqrt{3}\) et \(\tan \frac{4 \pi}{3} = \sqrt{3}\) (1pt), nous déterminons les expressions de :

\(\tan \left(x + \frac{2 \pi}{3}\right) = \frac{\tan x - \sqrt{3}}{1 + \sqrt{3} \tan x}\) (1pt)

\(\tan \left(x + \frac{4 \pi}{3}\right) = \frac{\tan x + \sqrt{3}}{1 - \sqrt{3} \tan x}\) (1pt)

donc

\(\begin{array}{ll}E(x) &= \tan x + \frac{\tan x - \sqrt{3}}{1 + \sqrt{3} \tan x} + \frac{\tan x + \sqrt{3}}{1 - \sqrt{3} \tan x} \\ & = \tan x + \frac{\left(\tan x - \sqrt{3}\right)\left(1-\sqrt{3} \tan x\right)+\left(\tan x + \sqrt{3}\right)\left(1+\sqrt {3} \tan x \right)}{1-3 \tan^{2} x} \end{array}\)

après simplification

\(\begin{array}{ll} E(x) &= \tan x + \frac{8 \tan x}{1-3 \tan^{2} x} \\ & = \frac{9 \tan x - 3 \tan^{3} x}{ 1 - 3 \tan^{2}x} \end{array}\)

donc \(E(x) = 3 \tan 3x\) (2pts)