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 .
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Lista de comentários
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 .