Propriété (Formule de Moivre) n ∈ N ∗ n \in \mathbb{N}^* n ∈ N ∗ x ∈ R x \in \mathbb{R} x ∈ R
( e i x ) n = ( cos ( x ) + i sin ( x ) ) n = cos ( n x ) + i sin ( n x ) (e^{ix})^n = (\cos(x) + i \sin(x))^n = \cos(nx) + i \sin(nx) ( e i x ) n = ( cos ( x ) + i sin ( x ) ) n = cos ( n x ) + i sin ( n x ) En effet : ( e i x ) n = e i n x \left(e^{ix}\right)^n = e^{inx} ( e i x ) n = e in x
Application x ∈ R x \in \mathbb{R} x ∈ R n ∈ N ∗ n \in \mathbb{N}^* n ∈ N ∗ cos ( n x ) \cos(nx) cos ( n x ) sin ( n x ) \sin(nx) sin ( n x ) cos ( x ) \cos(x) cos ( x ) sin ( x ) \sin(x) sin ( x )
( cos x + i sin x ) n = cos ( n x ) + i sin ( n x ) (\cos x + i \sin x)^n = \cos(nx) + i \sin(nx) ( cos x + i sin x ) n = cos ( n x ) + i sin ( n x ) On développe le membre de gauche, puis on en déduit :
{ cos ( n x ) = Re ( ( cos x + i sin x ) n ) sin ( n x ) = Im ( ( cos x + i sin x ) n ) \begin{cases}
\cos(nx) = \operatorname{Re}\left((\cos x + i \sin x)^n\right) \\
\sin(nx) = \operatorname{Im}\left((\cos x + i \sin x)^n\right)
\end{cases} { cos ( n x ) = Re ( ( cos x + i sin x ) n ) sin ( n x ) = Im ( ( cos x + i sin x ) n ) Exemple cos ( 2 x ) \cos(2x) cos ( 2 x )
( cos x + i sin x ) 2 = cos ( 2 x ) + i sin ( 2 x ) (\cos x + i \sin x)^2 = \cos(2x) + i \sin(2x) ( cos x + i sin x ) 2 = cos ( 2 x ) + i sin ( 2 x ) En développant :
( cos x + i sin x ) 2 = cos 2 x + 2 i cos x sin x − sin 2 x (\cos x + i \sin x)^2 = \cos^2 x + 2i \cos x \sin x - \sin^2 x ( cos x + i sin x ) 2 = cos 2 x + 2 i cos x sin x − sin 2 x Donc :
cos ( 2 x ) = cos 2 x − sin 2 x \cos(2x) = \cos^2 x - \sin^2 x cos ( 2 x ) = cos 2 x − sin 2 x Puisque sin 2 x = 1 − cos 2 x \sin^2 x = 1 - \cos^2 x sin 2 x = 1 − cos 2 x
cos ( 2 x ) = 2 cos 2 x − 1 \cos(2x) = 2 \cos^2 x - 1 cos ( 2 x ) = 2 cos 2 x − 1 Exercice cos ( 3 x ) \cos(3x) cos ( 3 x ) cos ( x ) \cos(x) cos ( x ) sin ( 3 x ) \sin(3x) sin ( 3 x ) sin ( x ) \sin(x) sin ( x )
Correction
Rappel :
( a + b ) 3 = a 3 + b 3 + 3 a b ( a + b ) (a + b)^3 = a^3 + b^3 + 3ab(a + b) ( a + b ) 3 = a 3 + b 3 + 3 ab ( a + b ) ( a − b ) 3 = a 3 − b 3 − 3 a b ( a − b ) (a - b)^3 = a^3 - b^3 - 3ab(a - b) ( a − b ) 3 = a 3 − b 3 − 3 ab ( a − b )
Calculons ( cos x + i sin x ) 3 = cos ( 3 x ) + i sin ( 3 x ) (\cos x + i \sin x)^3 = \cos(3x) + i \sin(3x) ( cos x + i sin x ) 3 = cos ( 3 x ) + i sin ( 3 x )
( cos x + i sin x ) 3 = cos 3 x + i 3 sin 3 x + 3 i cos x sin x ( cos x + i sin x ) = cos 3 x − i sin 3 x + 3 i cos 2 x sin x − 3 cos x sin 2 x = cos 3 x − 3 cos x sin 2 x + i ( − sin 3 x + 3 cos 2 x sin x ) \begin{aligned}
(\cos x + i \sin x)^3 &= \cos^3 x + i^3 \sin^3 x + 3i \cos x \sin x (\cos x + i \sin x) \\
&= \cos^3 x - i \sin^3 x + 3i \cos^2 x \sin x - 3 \cos x \sin^2 x \\
&= \cos^3 x - 3 \cos x \sin^2 x + i (-\sin^3 x + 3 \cos^2 x \sin x)
\end{aligned} ( cos x + i sin x ) 3 = cos 3 x + i 3 sin 3 x + 3 i cos x sin x ( cos x + i sin x ) = cos 3 x − i sin 3 x + 3 i cos 2 x sin x − 3 cos x sin 2 x = cos 3 x − 3 cos x sin 2 x + i ( − sin 3 x + 3 cos 2 x sin x ) D’où :
{ cos ( 3 x ) = cos 3 x − 3 cos x sin 2 x sin ( 3 x ) = − sin 3 x + 3 cos 2 x sin x \begin{cases}
\cos(3x) = \cos^3 x - 3 \cos x \sin^2 x \\
\sin(3x) = -\sin^3 x + 3 \cos^2 x \sin x
\end{cases} { cos ( 3 x ) = cos 3 x − 3 cos x sin 2 x sin ( 3 x ) = − sin 3 x + 3 cos 2 x sin x Or, on sait que cos 2 x + sin 2 x = 1 \cos^2 x + \sin^2 x = 1 cos 2 x + sin 2 x = 1
{ cos 2 x = 1 − sin 2 x sin 2 x = 1 − cos 2 x \begin{cases}
\cos^2 x = 1 - \sin^2 x \\
\sin^2 x = 1 - \cos^2 x
\end{cases} { cos 2 x = 1 − sin 2 x sin 2 x = 1 − cos 2 x Ainsi, on obtient :
{ cos ( 3 x ) = 4 cos 3 x − 3 cos x sin ( 3 x ) = 3 sin x − 4 sin 3 x \begin{cases}
\cos(3x) = 4 \cos^3 x - 3 \cos x \\
\sin(3x) = 3 \sin x - 4 \sin^3 x
\end{cases} { cos ( 3 x ) = 4 cos 3 x − 3 cos x sin ( 3 x ) = 3 sin x − 4 sin 3 x Activité
Soit θ \theta θ z = e i θ z = e^{i\theta} z = e i θ
Calculer z + z ˉ z + \bar{z} z + z ˉ cos ( θ ) \cos(\theta) cos ( θ ) θ \theta θ
Calculer z − z ˉ z - \bar{z} z − z ˉ sin ( θ ) \sin(\theta) sin ( θ ) θ \theta θ
Correction
Calculons z + z ˉ z + \bar{z} z + z ˉ
z + z ˉ = 2 Re ( z ) = 2 cos θ z + \bar{z} = 2 \operatorname{Re}(z) = 2 \cos \theta z + z ˉ = 2 Re ( z ) = 2 cos θ Donc :
cos ( θ ) = z + z ˉ 2 = e i θ + e − i θ 2 \cos(\theta) = \dfrac{z + \bar{z}}{2} = \dfrac{e^{i\theta} + e^{-i\theta}}{2} cos ( θ ) = 2 z + z ˉ = 2 e i θ + e − i θ
Calculons z − z ˉ z - \bar{z} z − z ˉ
z − z ˉ = 2 i Im ( z ) = 2 i sin θ z - \bar{z} = 2i \operatorname{Im}(z) = 2i \sin \theta z − z ˉ = 2 i Im ( z ) = 2 i sin θ Donc :
sin ( θ ) = z − z ˉ 2 i = e i θ − e − i θ 2 i \sin(\theta) = \dfrac{z - \bar{z}}{2i} = \dfrac{e^{i\theta} - e^{-i\theta}}{2i} sin ( θ ) = 2 i z − z ˉ = 2 i e i θ − e − i θ Propriété
Soit θ ∈ R \theta \in \mathbb{R} θ ∈ R
{ cos ( θ ) = e i θ + e − i θ 2 sin ( θ ) = e i θ − e − i θ 2 i \begin{cases}
\cos(\theta) = \dfrac{e^{i\theta} + e^{-i\theta}}{2}\\~\\
\sin(\theta) = \dfrac{e^{i\theta} - e^{-i\theta}}{2i}
\end{cases} ⎩ ⎨ ⎧ cos ( θ ) = 2 e i θ + e − i θ sin ( θ ) = 2 i e i θ − e − i θ Application : Linéarisation
La linéarisation consiste à transformer une fonction contenant des puissances de sinus et cosinus (cos n ( x ) \cos^n(x) cos n ( x ) sin m ( x ) \sin^m(x) sin m ( x ) cos ( a x ) \cos(ax) cos ( a x ) sin ( b x ) \sin(bx) sin ( b x ) a , b ∈ N a,b \in \mathbb{N} a , b ∈ N
Exemple
Linéarisons cos 2 ( x ) \cos^2(x) cos 2 ( x )
cos ( x ) = e i x + e − i x 2 ⇒ cos 2 ( x ) = ( e i x + e − i x 2 ) 2 \cos(x) = \dfrac{e^{ix} + e^{-ix}}{2}
\Rightarrow \cos^2(x) = \left(\dfrac{e^{ix} + e^{-ix}}{2}\right)^2 cos ( x ) = 2 e i x + e − i x ⇒ cos 2 ( x ) = ( 2 e i x + e − i x ) 2 Développement :
cos 2 ( x ) = ( e i x + e − i x 2 ) 2 = 1 4 ( e 2 i x + 2 + e − 2 i x ) = 1 2 ( e 2 i x + e − 2 i x 2 + 1 ) = 1 2 ( cos ( 2 x ) + 1 ) \begin{aligned}
\cos^2(x) &= \left( \dfrac{e^{ix} + e^{-ix}}{2} \right)^2
\\&= \dfrac{1}{4} \left( e^{2ix} + 2 + e^{-2ix} \right) \\
&= \dfrac{1}{2} \left( \dfrac{e^{2ix} + e^{-2ix}}{2} + 1 \right) \\
&= \dfrac{1}{2} \left( \cos(2x) + 1 \right)
\end{aligned} cos 2 ( x ) = ( 2 e i x + e − i x ) 2 = 4 1 ( e 2 i x + 2 + e − 2 i x ) = 2 1 ( 2 e 2 i x + e − 2 i x + 1 ) = 2 1 ( cos ( 2 x ) + 1 ) Exercice
Linéariser sin 3 ( x ) \sin^3(x) sin 3 ( x )
Linéariser cos ( x ) sin 2 ( x ) \cos(x)\sin^2(x) cos ( x ) sin 2 ( x )
Correction
1/ Linéarisation de sin 3 ( x ) \sin^3(x) sin 3 ( x )
Rappel :∙ ( a + b ) 3 = a 3 + b 3 + 3 a b ( a + b ) \bullet(a + b)^3 = a^3 + b^3 + 3ab(a + b) ∙ ( a + b ) 3 = a 3 + b 3 + 3 ab ( a + b )
∙ ( a − b ) 3 = a 3 − b 3 − 3 a b ( a − b ) \bullet(a - b)^3 = a^3 - b^3 - 3ab(a - b) ∙ ( a − b ) 3 = a 3 − b 3 − 3 ab ( a − b )
sin 3 ( x ) = ( e i x − e − i x 2 i ) 3 = e 3 i x − e − 3 i x − 3 e i x e − i x ( e i x − e − i x ) − 8 i = − 1 4 ( e 3 i x − e − 3 i x 2 i − 3 e i x − e − i x 2 i ) = − 1 4 ( sin ( 3 x ) − 3 sin ( x ) ) = − 1 4 sin ( 3 x ) + 3 4 sin ( x ) \begin{aligned}
\sin^3(x) &= \left( \dfrac{e^{ix} - e^{-ix}}{2i} \right)^3 \\
&= \dfrac{e^{3ix} - e^{-3ix} - 3e^{ix}e^{-ix}(e^{ix} - e^{-ix})}{-8i} \\
&= -\dfrac{1}{4} \left( \dfrac{e^{3ix} - e^{-3ix}}{2i} - 3 \dfrac{e^{ix} - e^{-ix}}{2i} \right) \\
&= -\dfrac{1}{4} \left( \sin(3x) - 3 \sin(x) \right) \\
&= -\dfrac{1}{4} \sin(3x) + \dfrac{3}{4} \sin(x)
\end{aligned} sin 3 ( x ) = ( 2 i e i x − e − i x ) 3 = − 8 i e 3 i x − e − 3 i x − 3 e i x e − i x ( e i x − e − i x ) = − 4 1 ( 2 i e 3 i x − e − 3 i x − 3 2 i e i x − e − i x ) = − 4 1 ( sin ( 3 x ) − 3 sin ( x ) ) = − 4 1 sin ( 3 x ) + 4 3 sin ( x ) 2/ Linéarisation de cos ( x ) sin 2 ( x ) \cos(x)\sin^2(x) cos ( x ) sin 2 ( x )
cos ( x ) sin 2 ( x ) = ( e i x + e − i x 2 ) ( e i x − e − i x 2 i ) 2 = ( e i x + e − i x 2 ) ( − ( e 2 i x − 2 + e − 2 i x ) 4 ) = 1 8 ( e 3 i x + 2 e i x + 1 + 2 e − i x + e − 3 i x ) = 1 4 ( cos ( 3 x ) + 2 cos ( x ) + 1 ) \begin{aligned}
\cos(x) \sin^2(x) &= \left( \dfrac{e^{ix} + e^{-ix}}{2} \right) \left( \dfrac{e^{ix} - e^{-ix}}{2i} \right)^2 \\
&= \left( \dfrac{e^{ix} + e^{-ix}}{2} \right) \left( -\dfrac{(e^{2ix} - 2 + e^{-2ix})}{4} \right) \\
&= \dfrac{1}{8} \left( e^{3ix} + 2e^{ix} + 1 + 2e^{-ix} + e^{-3ix} \right) \\
&= \dfrac{1}{4} \left( \cos(3x) + 2\cos(x) + 1 \right)
\end{aligned} cos ( x ) sin 2 ( x ) = ( 2 e i x + e − i x ) ( 2 i e i x − e − i x ) 2 = ( 2 e i x + e − i x ) ( − 4 ( e 2 i x − 2 + e − 2 i x ) ) = 8 1 ( e 3 i x + 2 e i x + 1 + 2 e − i x + e − 3 i x ) = 4 1 ( cos ( 3 x ) + 2 cos ( x ) + 1 )