Bonjour,
On a tan(2x) = 2 tan(x) / (1 - tan²(x))
D'où 2 tan(π/8) / (1 - tan²(π/8)) = 1
⇒ tan²(π/8) + 2 tan(π/8) + 1 = 2
⇒ (1 + tan(π/8))² = 2
⇒ 1 + tan(π/8) = √2 (car tan(π/8) > 0)
⇒ tan(π/8) = √2 - 1
1) D'autre part AE = AC = √2 donc BE = AE - AB = √2 - 1
D'où tan(BCE) = BE/BC = (√2 - 1)/1 = √2 - 1
Soit BCE = π/8
D'où DCE = π/8 + π/2 = 5π/8
2) DC.CE = DC . (CB + BE) = DC.BE = 1 * (√2 - 1) = √2 - 1
3) On a CE² = CB² + BE² (th. de Pythagore)
soit CE² = 1 + (√2 - 1)² = 1 + 2 + 1 - 2√2 = 4 - 2√2
D'autre part CD.CE = ||CD||.||CE||. cos (DCE) = 1 * √(4 - 2√2) . cos(5π/8)
D'où √(4 - 2√2) . cos(5π/8) = √2 - 1
⇒ cos(5π/8) = (√2 - 1) / (√2 . √(2 - √2)) = (√2 - 1) . √2 . √(2 - √2) / (2 (2 - √2))
⇒ cos(5π/8) = (2 - √2) . √(2 - √2) / (2 (2 - √2)) = √(2 - √2) / 2
sin²(5π/8) = 1 - cos²(5π/8) = 1 - (√(2 - √2) / 2)²
⇒ sin²(5π/8) = 1 - (2 - √2) / 4) = (2 + √2) / 4
⇒ sin(5π/8) = √(2 + √2) / 2
