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