Trigonométrie — Première Mathématiques

✅ Corrigé détaillé — Exercice 2

Partie A

Expression Formule d’angle associé Valeur exacte
a) cos(2π/3) cos(π − π/3) = −cos(π/3) −1/2
b) sin(5π/6) sin(π − π/6) = sin(π/6) 1/2
c) cos(−π/3) cos(−x) = cos(x) → cos(π/3) 1/2
d) sin(7π/6) sin(π + π/6) = −sin(π/6) −1/2
e) cos(5π/4) cos(π + π/4) = −cos(π/4) −√2/2
f) sin(11π/6) sin(2π − π/6) = sin(−π/6) = −sin(π/6) −1/2

Partie B

2a) cos(π−x) = −cos(x) = −1/3.
2b) sin(π+x) = −sin(x) = −(−2√2/3) = 2√2/3.
2c) cos(−x) = cos(x) = 1/3.
2d) sin(π/2 − x) = cos(x) = 1/3.

Partie C

3a) sin(100π + π/3) = sin(π/3 + 100π) = sin(π/3 + 50 × 2π) = sin(π/3) = √3/2.
(100π = 50 × 2π, donc on effectue 50 tours entiers).

3b) cos(−7π + π/4) = cos(−7π + π/4 + 4×2π) = cos(π + π/4) = −cos(π/4) = −√2/2.
(−7π + 8π = π, puis cos(π + π/4) = −cos(π/4).)

3c) sin(5π/3 + 8π) = sin(5π/3 + 4 × 2π) = sin(5π/3) = sin(2π − π/3) = sin(−π/3) = −sin(π/3) = −√3/2.

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