Linéarisations (1) - Corrigé

Énoncé

Linéariser les expressions suivantes.

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

Solution

Soit  \(x \in \mathbb{R}\) , on a :

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

donc \(A(x)=\dfrac{1}{2}\sin(3x)-\dfrac{1}{2}\sin(x)\) .

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

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

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

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

\(\begin{align*}D(x)=\sin(x)\sin(4x)= \left(\frac{\text e^{ix}-\text e^{-ix}}{2i}\right)\left(\frac{\text e^{4ix}-\text e^{-4ix}}{2i}\right)& = \frac{\text e^{ix}\text e^{4ix}-\text e^{ix}\text e^{-4ix}-\text e^{-ix}\text e^{4ix}+\text e^{-ix}\text e^{-4ix}}{-4}\\& = \frac{\text e^{5ix}-\text e^{-3ix}-\text e^{3ix}+\text e^{-5ix}}{4i}\\& = \frac{\left(\text e^{5ix}+\text e^{-5ix}\right)-\left(\text e^{3ix}+\text e^{-3ix}\right)}{-4}\\& = \frac{2\cos(5x)-2\cos(3x)}{-4}\\& = -\frac{1}{2}\sin(5x)+\frac{1}{2}\sin(3x)\end{align*}\)

donc \(F(x)=-\dfrac{1}{2}\sin(5x)+\dfrac{1}{2}\sin(3x)\) .