Soit `k in Z `

Représenter sur le cercle trigonométrique les points d'abscisses

` a = (pi)/3 + 2kpi `

` b = pi + 2kpi `

` c = (2kpi)/3 `

` d = -(pi)/2 + k(pi)/3 ; `

On a ` a = (pi)/3 + 2kpi `

alors abscisse principale du point `a` est `(pi)/3 `

`=> ( hat(vec(OI) , vec(OA))) = (pi)/3 `

On a ` b = pi + 2kpi `

alors abscisse principale du point `a` est `pi `

`=> ( hat(vec(OI) , vec(OB))) = pi `

Cherchons abscisse principale du point ` c = (2kpi)/3 `

` -pi < (2kpi)/3 <= pi `

` -3/2 < k <= 3/2 `

` k = { -1 , 0 , 1 } `

`=> c_1 = (-2pi)/3 ` , ` c_2 = 0 ` , `c_3 = (2pi)/3 `

` ( hat(vec(OI) , vec(OC_1))) = (-2pi)/3 `

` ( hat(vec(OI) , vec(OC_2))) = 0 `

` ( hat(vec(OI) , vec(OC_3))) = (2pi)/3 `

cherchons les abscisses ` d = -(pi)/2 + k(pi)/3 ; `

On a ` -pi < -(pi)/2 + k(pi)/3 <= pi `

` -(pi) +(pi)/2 < k(pi)/3 <= (pi)/2 + pi `

` -1/2 < k/3 <= 3/2 `

` -3/2 < k <= 9/2 `

`=> k in { -1, 0 , 1, 2 , 3 , 4 } `

`=> d_1 = -(pi)/2 -(pi)/3 = (-5pi)/6 => (hat(vec(OI) , vec(OD_1))) = (-5pi)/6 `

`=> d_2 = -(pi)/2 => (hat(vec(OI) , vec(OD_2))) = -(pi)/2 `

`=> d_3 = -(pi)/2 +(pi)/3 = -(pi)/6 => (hat(vec(OI) , vec(OD_3))) = -(pi)/6 `

`=> d_4= -(pi)/2 + (2pi)/3 = (pi)/6 => (hat(vec(OI) , vec(OD_4))) = (pi)/6 `

`=> d_5= -(pi)/2 + (3pi)/3 = (pi)/2 => (hat(vec(OI) , vec(OD_5))) = (pi)/2 `

`=> d_6= -(pi)/2 + (4pi)/3 = (5pi)/6 => (hat(vec(OI) , vec(OD_6))) = (5pi)/6 `