Bonjour pouvez vous m’aider s’il vous plaît

Bonjour ;

1.

On a pour tout x ∈ IR : cos(π - x) = cos(π + x) = - cos(x)

et cos(2π - x) = cos(- x) = cos(x) .

Si x = π/3 alors on a : 2π/3 = π - π/3 ;

donc : cos(2π/3) = cos(π - π/3) = - cos(π/3) .

De même on a : 4π/3 = π + π/3 ;

donc : cos(4π/3) = cos(π + π/3) = - cos(π/3) .

Enfin , on a : 5π/3 = 2π - π/3 ;

donc : cos(5π/3) = cos(π/3) .

Conclusion :

A = cos(π/3) - cos(2π/3) - cos(4π/3) + cos(5π/3)

= cos(π/3) - (- cos(π/3)) - (- cos(π/3)) + cos(π/3)

= cos(π/3) + cos(π/3) + cos(π/3) + cos(π/3)

= 4 cos(π/3) = 4 x 1/2 = 2 car cos(π/3) = 1/2 .

2.

B = cos(π/3 x cos(2π/3) x cos(4π/3) x cos(5π/3)

= cos(π/3) x (- cos(π/3)) x (- cos(π/3)) x cos(π/3)

= (cos(π/3))^4

= (1/2)^4

= 1/2^4

= 1/16 .

3.

On sait que pour tout x ∈ IR , sin(- x) = - sin(x) ; donc  :

C = sin(- 3π/4) - sin(- π/4) - sin(π/4) + sin(3π/4)

= - sin(3π/4) - (- sin(π/4)) - sin(π/4) + sin(3π/4)

= - sin(3π/4) + sin(π/4) - sin(π/4) + sin(3π/4)

= 0 .