📝 Exercices de la série▼
Exercice 1Exercice 2Exercice 3Exercice 4Exercice 5Exercice 6

Exercice 3

on pose : u=tanx2u=tan\dfrac x2 avec x≠(2k+1)πx\ne (2k+1)\pi où k∈Zk\in\Z,

Montrer que :

sin(x)1+cos(x)=u   et   1−cos(x)1+cos(x)=u2\dfrac{sin(x)}{1+cos(x)}=u~~~et ~~~\dfrac{1-cos(x)}{1+cos(x)}=u^2

Propriété : Soit a∈Ra\in\R avec a≠(2k+1)π, k∈Za\ne (2k+1)\pi,~k\in\Z on a :

  • cos(a)=1−tan2(x2)1+tan⁡2x2cos(a)=\dfrac{1-tan^2\left(\dfrac x2\right)}{1+\tan^2\dfrac x2}
  • sin(a)=2tan(x2)1+tan⁡2x2sin(a)=\dfrac{2tan\left(\dfrac x2\right)}{1+\tan^2\dfrac x2}
  • tan(a)=2tan(x2)1−tan⁡2x2tan(a)=\dfrac{2tan\left(\dfrac x2\right)}{1-\tan^2\dfrac x2}
sin(x)1+cos(x)=2u1+u21+1−u21+u2=2u1+u2+1−u2=u\begin{align*} \dfrac{sin(x)}{1+cos(x)} &=\dfrac{\dfrac{2u}{1+u^2}}{1+\dfrac{1-u^2}{1+u^2}} \\ &=\dfrac{2u}{1+u^2+1-u^2}\\ &=u \end{align*}
1−cos(x)1+cos(x)=1−1−u21+u21+1−u21+u2=1+u2−1+u21+u2+1−u2=u2\begin{align*} \dfrac{1-cos(x)}{1+cos(x)} &=\dfrac{1-\dfrac{1-u^2}{1+u^2}}{1+\dfrac{1-u^2}{1+u^2}} \\ &=\dfrac{1+u^2-1+u^2}{1+u^2+1-u^2} \\ &=u^2 \end{align*}