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C & E & D & TI
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Correction exercice
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### Correction exercice I

En utilisant les formules de Moivre, linéarisons les expressions suivantes :
a) $$A(x) = {\cos ^3}x$$
$$A(x) =$$ $${\left( {\frac{{Z + \overline Z }}{2}} \right)^3}$$ $$= \frac{1}{8}({Z^3} +$$ $$3{Z^2}\overline Z +$$ $$3Z{\overline Z ^2} +$$ $${\overline Z ^3})$$
$${Z^n} + {\overline Z ^n}$$ $$= 2\cos \left( {n\theta } \right)$$
$${Z^3} + {\overline Z ^3}$$ $$= 2\cos \left( {3\theta } \right)$$
$${Z^3} + {\overline Z ^3}$$ $$= 2\cos \left( {3\theta } \right)$$
$$Z + \overline Z =$$ $$2\cos \left( \theta \right)$$
$$A(\theta) =$$ $$\frac{1}{8}\left( {{Z^3} + {{\overline Z }^3}} \right) +$$ $$\frac{3}{8}Z\overline Z \left( {Z + \overline Z } \right)$$
$$A(\theta) =$$ $$\frac{1}{8}2\cos \left( {3\theta } \right)$$ $$+ \frac{3}{8}2\cos \left( \theta \right)$$ $$= \frac{1}{4}\cos \left( {3\theta } \right)$$ $$+ \frac{3}{4}\cos \left( \theta \right)$$

$$A(x) =$$ $$\frac{1}{4}\cos \left( {3x} \right)$$ $$+ \frac{3}{4}\cos \left( x \right)$$

b) $$B(x) = {\sin ^3}x$$
$$B(x) =$$ $${\sin ^3}x =$$ $${\left( {\frac{{Z + \overline Z }}{2}} \right)^3}$$
$${Z^n} - {\overline Z ^n}$$ $$= 2i\sin \left( {n\theta } \right)$$
$${Z^n} \times {\overline Z ^n} = 1$$
$$B(x) =$$ $${\left( {\frac{{Z - \overline Z }}{2}} \right)^3}$$ $$= \frac{1}{{{{\left( {2i} \right)}^3}}}$$ $$(2({Z^3} - {\overline Z ^3})$$ $$+ 3(Z - \overline Z ))$$
$$B(x) =$$ $$\frac{1}{{ - 8i}}(2i\sin 3x)$$ $$+ \frac{1}{{ - 8i}}(6i\sin x)$$

$$B(x) =$$ $$- \frac{1}{4}\sin 3x$$ $$- \frac{3}{4}\sin x$$

c) $$C(x) = {\cos ^5}\frac{x}{2}$$
$$C(x) =$$ $$\frac{1}{{32}}\left( {{Z^5} + {{\overline Z }^5}} \right) +$$ $$\frac{1}{{32}}5Z\overline Z \left( {{Z^3} + {{\overline Z }^3}} \right)$$ $$+ \frac{1}{{32}}10{Z^2}{\overline Z ^2}$$ $$\left( {Z + \overline Z } \right)$$
$$C(x) =$$ $$\frac{1}{{32}}2\cos 5\frac{x}{2}$$ $$+ \frac{1}{{32}}10\cos 3\frac{x}{2}$$ $$+ \frac{1}{{32}}20\cos \frac{x}{2}$$

$$C(x) =$$ $$\frac{1}{{16}}\cos \frac{{5x}}{2} +$$ $$\frac{5}{{16}}\cos \frac{{3x}}{2} +$$ $$\frac{{10}}{{16}}\cos \frac{x}{2}$$

d) $$D(x) =$$ $${\cos ^3}x{\sin ^3}x$$
$$D(x) =$$ $${\left( {\frac{{Z + \overline Z }}{2}} \right)^3}$$ $${\left( {\frac{{Z - \overline Z }}{{2i}}} \right)^3}$$ $$= \frac{1}{{ - 64i}}\left( {{Z^6} - {{\overline Z }^6}} \right)$$ $$- \frac{3}{{ - 64i}}{Z^2}{\overline Z ^2}$$ $$\left( {{Z^2} - {{\overline Z }^2}} \right)$$
$$D(x) =$$ $$\frac{1}{{ - 64i}}\left( {2i\sin 6x} \right)$$ $$- \frac{3}{{ - 64i}}\left( {2i\sin 2x} \right)$$

$$D(x) =$$ $$- \frac{1}{{32}}\sin 6x +$$ $$\frac{3}{{32}}\sin 2x$$

### Exercice II

En utilisant la formule d'Euler, linéarisons les expressions suivantes :
a) $$A(x) = {\cos ^3}x$$
$$A(x) =$$ $$\frac{1}{8}{\left( {{e^{ix}} + {e^{ - ix}}} \right)^3}$$ $$\frac{1}{8}\left( {{e^{3ix}} + {e^{ - 3ix}}} \right)$$ $$+ \frac{3}{8}\left( {{e^{ix}} + {e^{ - ix}}} \right)$$

• $${e^{nix}} + {e^{ - nix}}$$ $$= 2\cos nx$$
• $${e^{nix}} \times {e^{ - nix}}$$ $$= 1$$

$$A(x) =$$ $$\frac{1}{4}\left( {\cos 3x} \right)$$ $$+ \frac{3}{4}\left( {\cos x} \right)$$

b) $$B(x) = {\sin ^3}x$$
$$B(x) =$$ $$\sin 3x =$$ $${\left( {\frac{{{e^{ix}} - {e^{ - ix}}}}{{2i}}} \right)^3}$$ $$= \frac{1}{{ - 8i}}$$ $$= \frac{1}{{ - 8i}}$$ $$+ \frac{1}{{ - 8i}}$$ $$\left( {{e^{ix}} - {e^{ - ix}}} \right)$$

$$B(x) = -$$ $$\frac{1}{4}\sin 3x -$$ $$\frac{3}{4}\sin x$$

c) $$C(x) = {\cos ^5}\frac{x}{2}$$
d) $$D(x) =$$ $${\cos ^3}x{\sin ^3}x$$

### Correction exercice III

Linéarisons en utilisant :
a) Les formules d’Euler;
$${\cos ^2}x =$$ $${\left( {\frac{{{e^{ix}} + {e^{ - ix}}}}{2}} \right)^2}$$ $$= \frac{1}{4}({\left( {{e^{ix}}} \right)^2}$$ $$+ 2{e^{ix}}{e^{ - ix}}$$ $$+ {\left( {{e^{ - ix}}} \right)^2} =$$ $$\frac{1}{4}\left( {{e^{i2x}} + {e^{ - 2ix}}} \right)$$ $$+ \frac{1}{2}$$
b) Les formules usuelles trigonométriques
$$\cos (x + x) =$$ $$\cos (2x) =$$ $${\cos ^2}x - {\sin ^2}x$$ $$= {\cos ^2}x -$$ $$(1 - {\cos ^2}x)$$ $$= 2{\cos ^2}x - 1$$
$${\cos ^2}x =$$ $$\frac{1}{2}\cos 2x + \frac{1}{2}$$

Linéarisons en utilisant :
a) Les formules d’Euler;
$${\sin ^3}x$$
$${\sin ^3}x =$$ $${\left( {\frac{{{e^{ix}} - {e^{ - ix}}}}{{2i}}} \right)^3}$$ $$= - \frac{1}{4}$$ $$\left( {\frac{{{e^{i3x}} - {e^{ - i3x}}}}{{2i}}} \right)$$ $$- \frac{3}{4}$$ $$\left( {\frac{{{e^{ix}} - {e^{ - ix}}}}{{2i}}} \right)$$
$${\sin ^3}x =$$ $$- \frac{1}{4}\sin 3x$$ $$- \frac{1}{4}\sin 3x$$

### Correction exercice IV

Linéarisons $$2\left( {1 + {{\sin }^2}x} \right)$$ $${\cos ^2}x$$
$$2\left( {1 + {{\sin }^2}x} \right)$$ $${\cos ^2}x$$ $$= 2$$ $$\left( {1 + {{\left( {\frac{{{e^{ix}} - {e^{ - ix}}}}{{2i}}} \right)}^2}} \right)$$ $${\left( {\frac{{{e^{ix}} + {e^{ - ix}}}}{{2i}}} \right)^2}$$ $$= \frac{1}{8}$$ $$\left( {6 - {e^{2ix}} - {e^{ - 2ix}}} \right)$$ $$\left( {2 + {e^{2ix}} + {e^{ - 2ix}}} \right)$$
$$2\left( {1 + {{\sin }^2}x} \right)$$ $${\cos ^2}x =$$ $$\frac{1}{4}(5 + 4\cos 2x$$ $$- \cos 4x)$$