Bonjour,
1)
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2) Nous avons montré précédemment que pour tout , et
Ainsi S1 = 0 et S2 = 0, donc S1 + S2 = 0
Bonjour ;
On a : cos(2x) = cos²(x) - sin²(x) = 2cos²(x) - 1 ;
donc : cos(3x) = cos(2x + x) = cos(2x) cos(x) - sin(2x) sin(x)
= (2cos²(x) - 1) cos(x) - 2sin²(x) cos(x) = 2cos³(x) - cos(x) - 2(1 - cos²(x)) cos(x)
= 2cos³(x) - cos(x) - 2cos(x) + 2cos³(x) = 4cos³(x) - 3cos(x) ;
donc : cos(3x) + 3cos(x) = 4cos³(x) ;
donc : cos³(x) = (cos(3x) + 3cos(x))/4 .
sin(3x) = sin(2x + x) = sin(2x) cos(x) + cos(2x) sin(x)
= 2sin(x) cos²(x) + (2cos²(x) - 1 ) sin(x) = 2sin(x) cos²(x) + 2sin(x) cos²(x) - sin(x)
= 4sin(x) cos²(x) - sin(x) = 4sin(x) (1 - sin²(x)) - sin(x)
= 4sin(x) - 4sin³(x) - sin(x) = 3sin(x) - 4sin³(x) ;
donc : 4sin³(x) = 3sin(x) - sin(3x) ;
donc : sin³(x) = (3sin(x) - sin(3x))/4 .
2)
Comme on a demandé une déduction à partir de la question n° 1 , on n'utilisera pas la formule : cos(π + x) = cos(π - x) = - cos(x) .
On a :
cos³(π/12) = (cos(π/4) + 3cos(π/12))/4 ;
cos³(5π/12) = (cos(5π/4) + 3cos(5π/12))/4
= (cos(π + π/4) + 3cos(5π/12))/4 = (- cos(π/4) + 3cos(5π/12))/4 ;
cos³(7π/12) = (cos(7π/4) + 3cos(7π/12))/4
= (cos(2π - π/4) + 3cos(π - 5π/12))/4
= (cos(π/4) - 3cos(5π/12))/4 ;
cos³(11π/12) = (cos(11π/4 + 3cos(11π/12))/4
= (cos(3π - π/4) + 3cos(π - π/12))/4 = (- cos(π/4) - 3cos(π/12))/4 ;
donc : S1 = cos³(π/12) + cos³(5π/12) + cos³(7π/12) + cos³(11π/12)
= (cos(π/4) + 3cos(π/12) - cos(π/4) + 3cos(5π/12) - cos(π/4) - 3cos(π/12)
+ cos(π/4) - 3cos(5π/12))/4 = 0/4 = 0 .
De même pour S2 .
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Verified answer
Bonjour,
1)
------------------------------------------------------------------------------------------
2) Nous avons montré précédemment que pour tout , et
Ainsi S1 = 0 et S2 = 0, donc S1 + S2 = 0
Verified answer
Bonjour ;
1)
On a : cos(2x) = cos²(x) - sin²(x) = 2cos²(x) - 1 ;
donc : cos(3x) = cos(2x + x) = cos(2x) cos(x) - sin(2x) sin(x)
= (2cos²(x) - 1) cos(x) - 2sin²(x) cos(x) = 2cos³(x) - cos(x) - 2(1 - cos²(x)) cos(x)
= 2cos³(x) - cos(x) - 2cos(x) + 2cos³(x) = 4cos³(x) - 3cos(x) ;
donc : cos(3x) + 3cos(x) = 4cos³(x) ;
donc : cos³(x) = (cos(3x) + 3cos(x))/4 .
sin(3x) = sin(2x + x) = sin(2x) cos(x) + cos(2x) sin(x)
= 2sin(x) cos²(x) + (2cos²(x) - 1 ) sin(x) = 2sin(x) cos²(x) + 2sin(x) cos²(x) - sin(x)
= 4sin(x) cos²(x) - sin(x) = 4sin(x) (1 - sin²(x)) - sin(x)
= 4sin(x) - 4sin³(x) - sin(x) = 3sin(x) - 4sin³(x) ;
donc : 4sin³(x) = 3sin(x) - sin(3x) ;
donc : sin³(x) = (3sin(x) - sin(3x))/4 .
2)
Comme on a demandé une déduction à partir de la question n° 1 , on n'utilisera pas la formule : cos(π + x) = cos(π - x) = - cos(x) .
On a :
cos³(π/12) = (cos(π/4) + 3cos(π/12))/4 ;
cos³(5π/12) = (cos(5π/4) + 3cos(5π/12))/4
= (cos(π + π/4) + 3cos(5π/12))/4 = (- cos(π/4) + 3cos(5π/12))/4 ;
cos³(7π/12) = (cos(7π/4) + 3cos(7π/12))/4
= (cos(2π - π/4) + 3cos(π - 5π/12))/4
= (cos(π/4) - 3cos(5π/12))/4 ;
cos³(11π/12) = (cos(11π/4 + 3cos(11π/12))/4
= (cos(3π - π/4) + 3cos(π - π/12))/4 = (- cos(π/4) - 3cos(π/12))/4 ;
donc : S1 = cos³(π/12) + cos³(5π/12) + cos³(7π/12) + cos³(11π/12)
= (cos(π/4) + 3cos(π/12) - cos(π/4) + 3cos(5π/12) - cos(π/4) - 3cos(π/12)
+ cos(π/4) - 3cos(5π/12))/4 = 0/4 = 0 .
De même pour S2 .