Commit 4ac5bda2 authored by Claude Meny's avatar Claude Meny

Update cheatsheet.fr.md

parent b094088d
Pipeline #15657 canceled with stage
...@@ -614,30 +614,29 @@ $`\quad\boldsymbol{\mathbf{=\color{brown}{2\,A\cdot cos\Big(\dfrac{\varphi_1-\va ...@@ -614,30 +614,29 @@ $`\quad\boldsymbol{\mathbf{=\color{brown}{2\,A\cdot cos\Big(\dfrac{\varphi_1-\va
<br> <br>
$`\quad =A\cdot(c\alpha\,+\,i\,s\alpha) \cdot (c\varphi_1\,+\,i\,s\varphi_1\,+\,c\varphi_2\,+\,i\,s\varphi_2)`$ $`\quad =A\cdot(c\alpha\,+\,i\,s\alpha) \cdot (c\varphi_1\,+\,i\,s\varphi_1\,+\,c\varphi_2\,+\,i\,s\varphi_2)`$
<br> <br>
$`\quad =A\cdot(c\,\alpha\;+\;i\,s\,\alpha) \cdot \big[\,(c\varphi_1\,+\,c\varphi_2)`$ $`\quad =A\cdot(c\,\alpha\;+\;i\,s\,\alpha) \cdot \big[\,(c\varphi_1+c\varphi_2)`$$`\,+\, i\,(s\varphi_1+s\varphi_2)\,\big]`$
$`,+\, i\,(s\varphi_1\,+\,s\varphi_2)\,\big]`$
* L'onde réelle est la partie réelle de $`\underline{U}(x,t)`$ :
<br>
$`\mathbf{U(x,t)} = \mathscr{Re}[\,\underline{U}(x,t)\,]`$
<br>
$`\quad =A\cdot\big[\,c\alpha\,(c\varphi_1+ c\varphi_2) - s\alpha\,(s\varphi_1+ s\varphi_2)\,\big]`$
<br> <br>
$`\color{blue}{\scriptsize{\quad\left| \begin{align} &cos(a+b)=cos(a)\,cos(b)-sin(a)\,sin(b)\\ $`\color{blue}{\scriptsize{\quad\left| \begin{align} &cos(a+b)=cos(a)\,cos(b)-sin(a)\,sin(b)\\
&cos(a-b)=cos(a)\,cos(b)+-sin(a)\,sin(b)\end{align}\right.}}`$ &cos(a-b)=cos(a)\,cos(b)+-sin(a)\,sin(b)\end{align}\right.}}`$
$`\color{blue}{\scriptsize{ \quad\Longrightarrow cos(a+b)+cos(a-b)=2\,cos(a)\,cos(b)}}`$ $`\color{blue}{\scriptsize{ \quad\Longrightarrow cos(a+b)+cos(a-b)=2\,cos(a)\,cos(b)}}`$
$`\color{blue}{\scriptsize{ \quad\text{En posant } p=a+b \text{ et } q=a-b\;,}}`$ $`\color{blue}{\scriptsize{ \quad\text{En posant } p=a+b \text{ et } q=a-b\;,}}`$
$`\color{blue}{\scriptsize{\quad \text{nous obtenons } a = (p+q)\,/\,2 \text{ et } b = (p-q)\,/\,2.}}`$ $`\color{blue}{\scriptsize{\quad \text{nous obtenons } a = (p+q)\,/\,2 \text{ et } b = (p-q)\,/\,2.}}`$
$`\color{blue}{\scriptsize{ \quad\text{Nous retrouvons ainsi } cos(p) + cos(q) = 2\,cos\Big(\dfrac{p+q}{2}\Big)\,cos\Big(\dfrac{p-q}{2}\Big)}}`$ $`\color{blue}{\scriptsize{ \quad\text{Nous retrouvons ainsi }}}`$
$`\color{blue}{\scriptsize{ \quad\quad cos(p) + cos(q) = 2\,cos\Big(\dfrac{p+q}{2}\Big)\,cos\Big(\dfrac{p-q}{2}\Big)}}`$
<br> <br>
$`\quad \begin{align}=A\;\Big[\,&c\alpha\cdot 2\,cos\Big(\dfrac{\varphi_1+\varphi_2}{2}\Big)\,cos\Big(\dfrac{\varphi_1-\varphi_2}{2}\Big)\\
&- s\alpha\cdot 2\,sin\Big(\dfrac{\varphi_1+\varphi_2}{2}\Big)\,sin\Big(\dfrac{\varphi_1-\varphi_2}{2}\Big)\end{align}`$
(c\alpha\;+\;i\,s\,\alpha) \cdot \big(\,(c\,\varphi_1\;+\;c\,\varphi_2)\;+\; i\,(s\,\varphi_1\;+\;s\,\varphi_2)\,\big)\big]`$
\right\}\Longrightarrow\\
\quad\quad cos^2(a)=cos(a)cos(a)=\dfrac{1}{2}[cos(a+a)+cos(a-a)]\\
\quad\quad\quad\quad=\dfrac{1}{2}[1 + cos(2a)]}}`$
<br>
$`\quad =A\;\big[ (c\,\alpha\;+\;i\,s\,\alpha) \cdot \big(\,(c\,\varphi_1\;+\;c\,\varphi_2)\;+\; i\,(s\,\varphi_1\;+\;s\,\varphi_2)\,\big)\big]`$
i\,s\,\varphi_1\;+\;c\,\varphi_2\;+\;i\,s\,\varphi_2)\,]`$ i\,s\,\varphi_1\;+\;c\,\varphi_2\;+\;i\,s\,\varphi_2)\,]`$
......
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