Linéarisations (2) - Corrigé

Énoncé

Linéariser les expressions suivantes.

• \(A(x)=\sin^3(x)\)
  
• \(B(x)=\cos^4(x)\)
  
• \(C(x)=\sin^3(x)\cos(x)\)
  
• \(D(x)=\cos^3(x)\sin^2(x)\)

Solution

Soit \(x \in \mathbb{R}\) . On a :

\(\begin{align*}A(x)=\sin^3(x)& = \left(\frac{\text e^{ix}-\text e^{-ix}}{2i}\right)^3\\& = \frac{\text e^{3ix}-3\text e^{2ix}\text e^{-ix}+3\text e^{ix}\text e^{-2ix}-\text e^{-3ix}}{-8i}\\& = \frac{\text e^{3ix}-3\text e^{ix}+3\text e^{-ix}-\text e^{-3ix}}{-8i}\\& = \frac{\left(\text e^{3ix}-\text e^{-3ix}\right)-3\left(\text e^{ix}-\text e^{-ix}\right)}{-8i}\\& = \frac{2i\sin(3x)-3 \times 2i\sin(x)}{-8i}\\& = -\frac{1}{4}\sin(3x)+\frac{3}{4}\sin(x)\end{align*}\)
donc \(A(x)=-\dfrac{1}{4}\sin(3x)+\dfrac{3}{4}\sin(x)\) .

\(\begin{align*}B(x)=\cos^4(x)& = \left(\frac{\text e^{ix}+\text e^{-ix}}{2}\right)^4\\& = \frac{\text e ^{4ix}+4\text e^{3ix}\text e^{-ix}+6\text e^{2ix}\text e^{-2ix}+4\text e^{ix}\text e^{-3ix}+\text e^{-4ix}}{16}\\& = \frac{\text e^{4ix}+4\text e^{2ix}+6+4\text e^{-2ix}+\text e^{-4ix}}{16}\\& = \frac{\left(\text e^{4ix}+\text e^{-4ix}\right)+4\left(\text e^{2ix}+\text e^{-2ix}\right)+6}{16}\\& = \frac{2\cos(4x)+4 \times 2\cos(2x)+6}{16}\\& = \frac{1}{8}\cos(4x)+\frac{1}{2}\cos(2x)+\frac{3}{8}\end{align*}\)
donc \(B(x)=\dfrac{1}{8}\cos(4x)+\dfrac{1}{2}\cos(2x)+\dfrac{3}{8}\) .

\(\begin{align*}C(x)=\sin^3(x)\cos(x)& = \left(\frac{\text e^{ix}-\text e^{-ix}}{2i}\right)^3\left(\frac{\text e^{ix}+\text e^{-ix}}{2}\right)\\& = \frac{\text e^{3ix}-3\text e^{2ix}\text e^{-ix}+3\text e^{ix}\text e^{-2ix}-\text e^{-3ix}}{-8i} \times \frac{\text e^{ix}+\text e^{-ix}}{2}\\& = \frac{\text e^{3ix}-3\text e^{ix}+3\text e^{-ix}-\text e^{-3ix}}{-8i}\times \frac{\text e^{ix}+\text e^{-ix}}{2}\\& = \frac{ \text e^{4ix}-3\text e^{2ix}+3\text e^{0}-\text e^{-2ix}+ \text e^{2ix}-3\text e^{0}+3\text e^{-2ix}-\text e^{-4ix}}{-16i}\\& = \frac{\text e^{4ix}-\text e^{-4ix}-3\text e^{2ix}+3\text e^{-3ix}+\text e^{2ix}-\text e^{-2ix}}{-16i}\\& = \frac{\left(\text e^{4ix}-\text e^{-4ix}\right)-3\left(\text e^{2ix}-\text e^{-2ix}\right)+\left(\text e^{2ix}+\text e^{-2ix}\right)}{-16i}\\& = \frac{2i\sin(4x)- 3 \times 2i \sin(2x)+ 2i \sin(2x)}{-16i}\\& = \frac{2i\sin(4x)- 4i \sin(2x) }{-16i}\\& =- \frac{1}{8}\sin(4x)+ \frac{1}{4}\sin(2x)\end{align*}\)
donc \(C(x)=- \dfrac{1}{8}\sin(4x)+ \dfrac{1}{4}\sin(2x)\) .

\(\begin{align*}D(x)=\cos^3(x)\sin^2(x)& = \left(\frac{\text e^{ix}+\text e^{-ix}}{2}\right)^3\left(\frac{\text e^{ix}-\text e^{-ix}}{2i}\right)^2\\& = \frac{\text e^{3ix}+3\text e^{ix}+3\text e^{-ix}+\text e^{-3ix}}{8}\times \frac{\text e^{2ix}-2+\text e^{-2ix}}{-4}\\& = \frac{\text e^{5ix}-2\text e^{3ix}+\text e^{ix}+3\text e^{3ix}-6\text e^{ix}+3\text e^{-ix}+3\text e^{ix}-6\text e^{-ix}+3\text e^{-3ix}+\text e^{-ix}-2\text e^{-3ix}+\text e^{-5ix}}{-32}\\& = \frac{\text e^{5ix}+\text e^{-5ix}+\text e^{3ix}+\text e^{-3ix}-2\text e^{ix}-2\text e^{-ix}}{-32}\\& = \frac{\left(\text e^{5ix}+\text e^{-5ix}\right)+\left(\text e^{3ix}+\text e^{-3ix}\right)-2\left(\text e^{ix}+\text e^{-ix}\right)}{-32}\\& = \frac{2\cos(5x)+2\cos(3x)-2 \times 2\cos(x)}{-32}\\& = -\frac{1}{16}\cos(5x)-\frac{1}{16}\cos(3x)+\frac{1}{8}\cos(x)\end{align*}\)
donc \(D(x)=-\dfrac{1}{16}\cos(5x)-\dfrac{1}{16}\cos(3x)+\dfrac{1}{8}\cos(x)\) .