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misssarita1
@misssarita1
January 2021
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Bonsoir s'il vous plait aidez moi à calculer :tang1°tang2°tang3°tang4°tang5°×...tang88°tang89°
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Geijutsu
Verified answer
Bonsoir,
On sait que pour tout réel x, tan(x) = sin(x)/cos(x)
Donc pour tout réel x compris entre 1 et 89, tan(x°)*tan(90-x°) = 1
Donc :
tan(1°)*tan(89°) = 1
tan(2°)*tan(88°) = 1
tan(3°)*tan(88°) = 1
.
.
.
tan(49°)*tan(51°) = 1
Donc tan(1°)*tan(2°)*...*tan(89°) = tan(50°) ≈ 1.1918
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Lista de comentários
Verified answer
Bonsoir,On sait que pour tout réel x, tan(x) = sin(x)/cos(x)
Donc pour tout réel x compris entre 1 et 89, tan(x°)*tan(90-x°) = 1
Donc :
tan(1°)*tan(89°) = 1
tan(2°)*tan(88°) = 1
tan(3°)*tan(88°) = 1
.
.
.
tan(49°)*tan(51°) = 1
Donc tan(1°)*tan(2°)*...*tan(89°) = tan(50°) ≈ 1.1918