Trigonométrie circulaire - Trigonométrie hyperbolique

Comme \(( \pi/4 - \textrm{Arctan }1/5) \in [ 0; \pi/4]\), les valeurs de x cherchées seront telles que : \(0 ≤ \textrm{Arctan }x + \textrm{Arctan }4x ≤ \pi/4\).

Donc par un calcul formel, prenons la tangente des deux membres :

\(\begin{array}{lll}\tan(\textrm{Arctan }x + \textrm{Arctan }4x) &= \frac{\tan(\textrm{Arctan }x) + \tan(\textrm{Arctan }4x)}{1-\tan(\textrm{Arctan }x)\tan(\textrm{Arctan }4x)}\\& = \frac{x+4x}{1-(x)(4x)}\end{array}\)

\(\tan(\textrm{Arctan }x + \textrm{Arctan }4x) = \frac{5x}{1-4x^{2}}\) (2pts)

\(\begin{array}{lll}\tan\left(\frac{\pi}{4} - \textrm{Arctan }\frac{1}{5}\right) &= \frac{\tan \frac{\pi}{4} - \tan\left(\textrm{Arctan }\frac{1}{5}\right)}{1 + \left(\tan \frac{\pi}{4}\right) \tan \left(\textrm{Arctan }\frac{1}{5}\right)} \\ &= \frac{1-\tfrac{1}{5}}{1+\tfrac{1}{5}} \end{array}\)

\(\tan \left(\frac{\pi}{4} - \textrm{Arctan} \frac{1}{5}\right) = \frac{2}{3}\) (1pt)

d'où \(\frac{5x}{1-4x^{2}} = \frac{2}{3}\)

\(\Leftrightarrow 2 - 8 x^{2} = 15x \Leftrightarrow 8 x^{2} + 15 x - 2 = 0\) (1pt)

L'équation qui admet pour racines :

\(\Delta = (15)^{2} + 4 \times 8 \times 2 = (17)^{2}\)

\(x = \frac{-15 \pm 17}{16} = \left\{\begin{array}{llll} x_{1} &=& -2 \\&\textrm{et}&\\x_{2} &=& 1/8\end{array}\right.\) (1pt)

Seule la solution \(x_{2} = 1/8\) vérifie la double inégalité \(0 \le \textrm{Arctan }x+ \textrm{Arctan }4x \le \pi/4.\) (2pts)