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Essaie de corriger sur la trigos
Exercice 2 :
1-Montrons que : ∀x∈R, cos^6 x+sin^6 x=(1/8)(5+3cos4x)
Soit x∈R, On a :
cos^6 x+sin^6 x=(cos^2 x)^3 +(sin^2 x)^3
=(cos^2 x+sin^2 x)^3 −3(sin^2 x)(cos^2 x)(cos^2 x+sin^2 x)
=1−3×(1/4)×2^2 .sin^2 x.cos^2 x
=1−(3/4)(sin^2 2x)
= 1−(3/8)(2sin^2 2x+1−1)
=1−(3/8)(−cos4x+1)
⇒cos^6 x+sin^6 x=(1/8)(5+3cos4x).
2.Re^ solvons ]−π;π] l′e^ quation (E),puis repre^ sentons dans le cercle trigos les solutions:
On a :
(E): cos^6 x+sin^6 x=(3/8)((√3)sin4x−(8/3))
D′apres ce qui pre^ ce^ de,
cos^6 x+sin^6 x=(3/8)((√3)sin4x−(8/3)) ⇒(1/8)(5+3cos4x)=(3/8)((√3)sin4x−(8/3))
⇒cos4x−(√3)sin4x=1
⇒cos ((π/3)+4x)=cos((π/3))
⇒ { ((x_1 =((kπ)/2))),((x_2 =−(π/6)+((kπ)/2))) :} (k∈Z)
determinant ((k,(−2),(−1),0,1,2),(x_1 ,(−π),(−(π/2)),0,(π/2),π),(x_2 ,(−((7π)/6)),(−((2π)/3)),(−(π/6)),(π/3),((5π)/6)))
determinant ((( S_(]−π;π]) ={−(π/2);−((2π)/3);−(π/6);0;(π/3);(π/2);((5π)/6);π} )))
Exercice 18
Soient x,y∈I=[0;(π/2)]/sinx=(((√6)−(√2))/4) et cosy=((√3)/2)
1.Calcule basic (anti 0) :
((((√6)−(√2))/4))^2 =((2−(√3))/4)
2.On a :
cos^2 x+sin^2 y=1⇒cosx=(√(1−sin^2 x)) car x∈I
ie cosx=(((√6)+(√2))/4)
3.De me^ me siny=(√(1−cos^2 y)) ie siny=(1/2)
il est evident que y=(π/6)
4. On a cos(x+y)=cosxcosy−sinxsiny
=(((√6)+(√2))/4) .((√3)/2)−(((√6)−(√2))/4).(1/2)
donc cos(x+y)=((√2)/2)
on a alors cos(x+y)=((√2)/2) ⇒x+y=(π/4) ie x=(π/(12))
5.Soit x∈R , sin^2 xcos^3 x=cosx(sin^2 x(1−sin^2 x))
ie sin^2 xcos^3 x=cosx(sin^2 x−sin^4 x)
donc ∀x∈R , sin^2 xcos^3 x=cosx(sin^2 x(1−sin^2 x))
Exercice 14
1 Re^ solvons dans R l′e^ quation: 2t^2 +(√3)t−3=0
Δ=27⇒(√Δ)=3(√3)
donc { ((t_1 =((−(√3)+3(√3))/(2(2))) )),((t_2 =((−(√3)−3(√3))/(2(2))) )) :} ⇒ { ((t_1 =((√3)/2))),((t_2 =−(√3))) :}
d′ou^ determinant ((( S_R ={−(√3);((√3)/2)})))
2.De^ terminons (r;ϕ)∈R_+ ×]0;2π[ /(√3)cosx+sin x=rcos(x−ϕ) :
(√3)cosx+sin x=2(((√3)/2)cosx+(1/2)sinx)
=2(cos((π/6))cosx+sin((π/6))sinx)
⇒(√3)cosx+sin x=2cos(x−(π/6))
donc (r;ϕ)=(2;(π/6))
3.Re^ solvons dans ]0;2π] l′e^ quation (E) :
(E):(2sin^2 x+(√3)sinx−3)( (√3)cosx+sin x−(√2))=0⇔2sin^2 x+(√3)sinx−3=0 ou (√3)cosx+sin x−(√2)=0
• 2sin^2 x+(√3)sinx−3=0
Posons t=sinx⇒2t^2 +(√3)t−3=0
i.e { ((t=((√3)/2))),((t=−(√3))) :}⇒sinx=((√3)/2)
=sin((π/3))
donc { ((x_1 =(π/3)+2kπ )),((x_2 =(π/3)+2(k−1)π)) :}(k∈Z)
determinant ((k,(−2),(−1),0,1,2),(x_1 ,(−((11π)/3)),(−((2π)/3)),(π/3),((8π)/3),((13π)/3)),(x_2 ,(−((14π)/3)),(−((8π)/3)),(−((2π)/3)),((4π)/3),((10π)/3)))
determinant (((S_(]0;2π]) ={(π/3);((4π)/3)})))
• (√3)cosx+sin x−(√2)=0 ⇒cos(x−(π/6))=((√2)/2)
=cos((π/4))
donc { ((x_1 =((5π)/(12))+2kπ)),((x_2 =−(π/(12))+2kπ)) :} (k∈Z)
determinant ((k,(−2),(−1),0,1,2),(x_1 ,(−((43π)/(12))),(−((19π)/(12))),((5π)/(12)),((29π)/(12)),((53π)/(12))),(x_2 ,(−((49π)/(12))),(−((25π)/(12))),(−(π/(12))),((23π)/(12)),((47π)/(12))))
determinant (((S_(]0;2π]) ={((5π)/(12));((23π)/(12))})))
Exercice 39
Soit x∈]0;2π]
1.Montrons que A(x)=4cos2x
On a:
A(x)=((cos3x)/(cosx))+((sin3x)/(sinx))
= ((sinx(cosx.cos2x−sin2x.sinx)+cosx(sin2x.cosx+cos2x.sinx))/(cosx.sinx))
=((sinx.cosx.cos2x−sin^2 x.sin2x+cos^2 x.sin2x+cos2x.sinx.cosx)/(cosx.sinx))
=((sin2x cos2x−2sin2x.sin^2 x+2cos^2 x.sin2x+cos2x.sin2x)/(sin2x))
=cos2x−2sin^2 x+2cos^2 x+cos2x
=2cos2x+2(cos^2 x−sin^2 x)
⇒A(x)=4cos2x
2.Re^ solvons A(x)=B(x)
A(x)=B(x)⇔4cos2x=4(1−(√3)sin2x)
⇔cos2x+(√3)sin2x=1
⇔cos(2x−(π/3))=(1/2)
=cos ((π/3))
⇒ { ((x_1 =(π/3)+kπ)),((x_2 =kπ)) :} (k∈Z)
determinant ((k,(−2),(−1),0,1,2),(x_1 ,(−((5π)/3)),(−((2π)/3)),(π/3),((4π)/3),((7π)/3)),(x_2 ,(−2π),(−π),0,π,(2π)))
determinant (((S_(]0;2π[) ={(π/3);π;((4π)/3)})))
Exercice 34
1. Montrons que (E_1 ) et (E_2 ) sont e^ quivalent
Soit x∈[0;2π], on a :
(E_1 ):sinx.cosx+cos^2 x=cos2x⇒sinx.cosx+cos^2 x=cos^2 x−sin^2 x
⇒sinx.cosx+sin^2 x=0
donc (E_1 )⇒(E_2 ).
(E_2 ):sinx.cosx+sin^2 x=0⇒sinx.cosx=−sin^2 x
⇒ sinx.cosx+cos^2 x=cos^2 x−sin^2 x
donc (E_2 )⇒(E_1 ).
Conclusion: (E_1 )⇔(E_2 )
2.Re^ solvons dans [0;2π] l′e^ quation (E_1 ) :
On a :
(E_1 ):sinx.cosx+cos^2 x=cos2x ⇔ sinx(cosx+sinx)=0
⇔ sinx.cos(x−(π/4))=0
⇔ sinx=0 ou cos(x−(π/4))=0
•sinx=0⇒x_1 =2kπ (k∈Z)
•cos(x−(π/4))=0⇒ { ((x_2 =((3π)/4)+2kπ)),((x_3 =−(π/4)+2kπ)) :} (k∈Z)
determinant ((k,0,1,2),(x_1 ,0,(2π),(4π)),(x_2 ,(((3π)/4) ),((11π)/4),((19π)/4)),(x_3 ,(−(π/4)),((7π)/4),((15π)/4)))
determinant (((S_(]0;2π[) ={0;((3π)/4);((7π)/4);((11π)/4);2π})))
Exercice 27
1.Re^ solvons dans [−π;π[ l′e^ quation (E): cos^2 2x=(1/2)
On a :
cos^2 2x=(1/2) ⇔cos2x=((√2)/2) ou cos2x=−((√2)/2)
⇔ { ((x_1 =(π/8)+kπ)),((x_2 =−(π/8)+kπ)) :} (k∈Z) ou { ((x_3 =((3π)/8)+kπ)),((x_4 =−((3π)/8)+kπ)) :}(k∈Z)
determinant ((k,(−2),(−1),0,1,2),(x_1 ,(−((15π)/8)),(−((7π)/8)),(π/8),((9π)/8),((17π)/8)),(x_2 ,(−((17π)/8)),(−((9π)/8)),(−(π/8)),((7π)/8),((15π)/8)),(x_3 ,(−((13π)/8)),(−((5π)/8)),((3π)/8),((11π)/8),((19π)/8)),(x_4 ,(−((19π)/8)),(−((11π)/8)),(−((3π)/8)),((5π)/8),((13π)/8)))
determinant ((( S_([−π;π[) ={−((7π)/8);−((5π)/8);−((3π)/8);−(π/8);(π/8);((3π)/8);((5π)/8);((7π)/8)} )))
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