AbderraoufSlamani
Bonsoir !! ABC est un triangle rectangle en A alors.. BAC = 90°
ABC = x ≠ 0 on note ACB = y ≠ 0 180 = x + y + 90 90 = x + y 0 < x < 90 sin 0 < sin x < sin 90 0 < sin x < 1
(cos x)² + (sin x)² = 1 (je ne sais pas comment monter)
tan x = AC/AB sin x = AC/CB cos x = AB/CB (sin x)/(cos x) = (AC/CB)/(AB/CB) (sin x)/(cos x) = (AC/CB) × (CB/AB) (sin x)/(cos x) = (AC × CB)/(CB × AB) (sin x)/(cos x) = AC/AB = tan x donc tan x = (sin x)/(cos x)
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ABC est un triangle rectangle en A alors..
BAC = 90°
ABC = x ≠ 0
on note ACB = y ≠ 0
180 = x + y + 90
90 = x + y
0 < x < 90
sin 0 < sin x < sin 90
0 < sin x < 1
(cos x)² + (sin x)² = 1 (je ne sais pas comment monter)
tan x = AC/AB
sin x = AC/CB
cos x = AB/CB
(sin x)/(cos x) = (AC/CB)/(AB/CB)
(sin x)/(cos x) = (AC/CB) × (CB/AB)
(sin x)/(cos x) = (AC × CB)/(CB × AB)
(sin x)/(cos x) = AC/AB = tan x
donc tan x = (sin x)/(cos x)