1 b) Calculer `cos((5pi)/(12) )` ;`sin((5pi)/(12)) ` ; `tan((5pi)/(12) )`
Rappels
Calculons `cos((5pi)/(12))`
comme ` (5pi)/(12)= (pi)/4 +(pi)/6`
`=> cos((5pi)/(12)) = cos((pi)/4 +(pi)/6)`
` = cos((pi)/4) cos((pi)/6) - sin((pi)/4) sin((pi)/6) `
` = (sqrt(2))/2xx(sqrt(3))/2 - (sqrt(2))/2 xx1/2 `
` = (sqrt(3xx2))/4 - (sqrt(2))/4 `
` = (sqrt(6) -sqrt(2))/4 `
alors
Calculons `sin((5pi)/(12))`
on a ` sin((5pi)/(12)) = sin((pi)/4 +(pi)/6)`
` = sin((pi)/4) cos((pi)/6) + sin((pi)/6) cos((pi)/4) `
` = (sqrt(2))/2xx(sqrt(3))/2 + 1/2xx(sqrt(2))/2 `
` = (sqrt(2xx3))/4 + (sqrt(2))/4 `
` = (sqrt(6) + sqrt(2))/4 `
Calculons `tan((5pi)/(12))` : Méthode 1
on a ` tan((5pi)/(12)) = (sin((5pi)/(12)))/(cos((5pi)/(12)))`
` = [ (sqrt(6) + sqrt(2))/4]/[ (sqrt(6) - sqrt(2))/4] `
` = (sqrt(6) +sqrt(2))/(sqrt(6) -sqrt(2)) `
` = (sqrt(6) +sqrt(2))^2/( 6 -2) `
` = ( 8 +2sqrt(12))/4 `
` = (8 + 4sqrt(3))/4 `
` = 2 +sqrt(3) `
alors
Calculons `tan((5pi)/(12))` : Méthode 2
on a `tan((5pi)/(12)) = (tan((pi)/4) + tan((pi)/6))/(1 -tan((pi)/4)tan((pi)/6) `
` = (1 + (sqrt(3))/3) /(1 - (sqrt(3))/3) `
` = [(3 + sqrt(3))/3]/[(3 - sqrt(3))/3]`
` = (3+sqrt(3))/(3-sqrt(3))`
` = (3+sqrt(3))^2/( 9 -3) `
` = (12+6sqrt(3))/6 `
` = 2 + sqrt(3) `
alors