Articles
Sciences et Sociétés
Mathématiques Algèbre

Bien savoir se rappeler des formules suivantes:

• cos (π - x ) = - cos x
• sin (π/2 - x) = + cos x
• sin (π - x) = + sin x
• cos(π/2 + x) = - sin x
• cos (π/2 - x) = + sin x
• cos (π + x) = - cos x
• cos (π - x) = - cos x
• sin (π + x) = - sin x
• cos (- x) = cos x
• sin (- x) = - sin x
• cos²x + sin ² = 1

Dans le deux exercices suivants, seules les formules pointées en rouge sont utilisées.


Exercice 1:

A = cos²(π/10) + cos²(2π/5) + cos²(3π/5) + cos²(9π/10)

On construit les relations suivantes:

cos(2π/5) = cos (π - 3π/5) = - cos (3π/5)      (1)

cos (3π/5) = sin (π/2 - 3π/5) = sin (-π/10)
= - sin (π/10)      (2)

(1) et (2) donnent:

cos²(2π/5) = cos²(3π/5) = sin²(π/10) (3)

cos(9π/10) = cos (π - π/10) = - cos (π/10)     (4)

Avec les relations (1), (2) et (4), on aura :

A = cos²(π/10) + cos²(2π/5) + cos²(2π/5) + cos²(π/10)
= 2 cos²(π/10) + 2 cos²(2π/5)

Avec la relation (3):

A = 2 cos²(π/10) + 2 sin²(π/10)
= 2 [cos²(π/10) + sin²(π/10)] = 2 x 1 = 2

A = 2


Exercice 2:

B = sin²(π/12) + sin²(3π/12) + sin²(5π/12) + sin²(7π/12) + sin²(9π/12) + sin²(11π/12)

On construit les relations suivantes:

• 5π/12 = 6π/12 - π/12 = π/2 - π/12

→ sin(5π/12) = sin(π/2 - π/12 ) = cos (π/12)      (1)

• 7π/12 = 6π/12 + π/12 = π/2 + π/12

→ sin(7π/12) = sin(π/2 + π/12 ) = cos (π/12)      (2)

• 9π/12 = 12π/12 - 3π/12 = π - π/4

→ sin(9π/12) = sin(π - 3π/12) = sin(π - π/4) = + sin (π/4) = √2/2      (3)

• 11π/12 = 12π/12 - π/12 = π - π/12

→ sin(11π/12) = sin(π - π/12) = + sin (π/12)      (4)

Il vient donc:

B = sin²(π/12) + sin²(3π/12) + cos²(π/12)
+ sin²(7π/12) + sin²(9π/12) + sin²(11π/12)

= sin²(π/12) + sin²(3π/12) + cos²(π/12)
+ cos²(π/12) + sin²(π/4) + sin²(π/12)

= 2 sin²(π/12) + 2 sin²(π/4) + 2 cos²(π/12)

= 2 [(sin²(π/12) + cos²(π/12)] + 2 sin²(π/4)
= 2 x 1 + 2 (√2/2)² = 2 + 2 x 2/4 = 2 + 1 = 3

B = 3

-- Abdurrazzak Ajaja
janvier 2023