Bonsoir,
Soit x un réel tel que |sin(x)cos(x)| = 1/4 et |tan(x)| < 1
1-
• (|sin(x)|+|cos(x)|)² = |sin(x)|²+2|sin(x)||cos(x)|+|cos(x)|² = sin²(x)+2|sin(x)cos(x)|+cos²(x) = 2(1/4)+sin²(x)+cos²(x) = 2(1/4)+1 = (1/2)+1 = 3/2
• (|sin(x)|-|cos(x)|)² = |sin(x)|²-2|sin(x)||cos(x)|+|cos(x)|² = sin²(x)-2|sin(x)cos(x)|+cos²(x) = -2(1/4)+sin²(x)+cos²(x) = 2(1/4)+1 = -(1/2)+1 = 1/2
2-
• (|sin(x)|+|cos(x)|)² = 3/2 ⇒ |sin(x)|+|cos(x)| = √(3/2) = (√6)/2 ou |sin(x)|+|cos(x)| = -√(3/2) = -(√6)/2
Or |sin(x)| ≥ 0 et |cos(x)| ≥ 0, d'où |sin(x)|+|cos(x)| ≥ 0
Donc |sin(x)|+|cos(x)| = (√6)/2
• (|sin(x)|-|cos(x)|)² = 1/2 ⇒ |sin(x)|+|cos(x)| = √(1/2) = (√2)/2 ou |sin(x)|+|cos(x)| = -√(1/2) = -(√2)/2
Or |tan(x)|< 1, d'où |sin(x)|/|cos(x)| < 1, d'où |sin(x)| < |cos(x)| car |cos(x)| ≥ 0, d'où |sin(x)|-|cos(x)| < 0
Donc |sin(x)|-|cos(x)| = -(√2)/2
3-
• On a alors le système suivant :
[tex] \left \{ {{(E):|sin(x)|+|cos(x)|=\frac{\sqrt{6}}{2}} \atop {(F):|sin(x)|-|cos(x)|=-\frac{\sqrt{2}}{2}}} \right.[/tex]
D'où (E)+(F) : |sin(x)|+|cos(x)|+|sin(x)|-|cos(x)| = ((√6)/2)-((√2)/2)
D'où 2|sin(x)| = (√6-√2)/2
Donc |sin(x)| = ((√6-√2)/2)/2 = (√6-√2)/4
• Ainsi, on obtient l'équation (E) : ((√6-√2)/4)+|cos(x)| = (√6)/2
Donc |cos(x)| = ((√6)/2)-((√6-√2)/4) = ((2√6)/4)-((√6-√2)/4) = (2√6-√6+√2)/4 = (√6+√2)/4
• Donc |tan(x)| = |sin(x)|/|cos(x)| = ((√6-√2)/4)/((√6+√2)/4) = (√6-√2)/(√6+√2) = 2-√3
