Chapitre VII Trigonométrie

\(\begin{cases}x= \frac{\pi}{6}+2k\pi\\x=-\frac{\pi}{6}+2k\pi\end{cases} k\in \mathbb{Z}\)

\(sin x=-0,5\)

\(\iff sin x=sin(-\frac{\pi}{6})\)

\(\iff \begin{cases}x=-\frac{\pi}{6}+2k\pi \\ x=\pi-(-\frac{\pi}{6})+2k\pi\end{cases}\) \(k\in \mathbb{Z}\)

\(\iff \begin{cases}x=-\frac{\pi}{6}+2k\pi \\ x=\pi+\frac{\pi}{6}+2k\pi\end{cases}\) \(k\in \mathbb{Z}\)

\(\iff \begin{cases}x=-\frac{\pi}{6}+2k\pi \\ x=\frac{6\pi}{6}+\frac{\pi}{6}+2k\pi\end{cases}\) \(k\in \mathbb{Z}\)

\(\iff \begin{cases}x=-\frac{\pi}{6}+2k\pi \\ x=\frac{7\pi}{6}+2k\pi\end{cases}\) \(k\in \mathbb{Z}\)

\(sin(3x)=sin(\frac{\pi}{2})\)

\(\iff \begin{cases}3x=\frac{\pi}{2}+2k\pi\\3x=\pi-\frac{\pi}{2}+2k\pi\end{cases}\) \(k \in \mathbb{Z}\)

\(\iff \begin{cases}x=\frac{\pi}{6}+\frac{2k\pi}{3}\\3x=\frac{\pi}{2}+2k\pi\end{cases}\) \(k \in \mathbb{Z}\)

\(\iff \begin{cases}x=\frac{\pi}{6}+\frac{2k\pi}{3}\\x=\frac{\pi}{6}+\frac{2k\pi}{3}\end{cases}\) \(k \in \mathbb{Z}\)

\(\iff x=\frac{\pi}{6}+\frac{2k\pi}{3}\) \(k \in \mathbb{Z}\)

Les mesures principales correspondantes sont :

• Pour \(k=0\): \(\frac{\pi}{6}\)

• Pour \(k=1\):\(\frac{\pi}{6}+\frac{2\pi}{3}=\frac{\pi}{6}+\frac{4\pi}{6}=\frac{5\pi}{6}\)

• Pour \(k=2\):

  \(\frac{\pi}{6}+\frac{4\pi}{3}=\frac{\pi}{6}+\frac{8\pi}{6}=\frac{9\pi}{6}-2\pi\)

  \(=\frac{9\pi}{6}-\frac{12\pi}{6}=\frac{-3\pi}{6}=\frac{-\pi}{2}\)

• Pour \(k=3\):\(\frac{\pi}{6}+\frac{6\pi}{3}=\frac{\pi}{6}+2\pi=\frac{\pi}{6}\)

  On retrouve la valeur pour \(k=0\), inutile de prendre des valeurs de \(k\) supérieures car on ajoute  \(\frac{2\pi}{3}\)

  entre deux angles successifs.

• Pour \(k=-1\): \(\frac{\pi}{6}+\frac{-2\pi}{3}=\frac{\pi}{6}-\frac{4\pi}{6}=\frac{-3\pi}{6}=-\frac{\pi}{2}\)

  On retrouve la valeur pour \(k=2\), inutile de prendre des valeurs de \(k\) inférieures car on soustrait  \(\frac{2\pi}{3}\)

  entre deux angles successifs.

\(cos(\frac{\pi}{8})=\frac{\sqrt{2+\sqrt{2}}}{2}\)

\((cos (\frac{\pi}{8}))^2+(sin (\frac{\pi}{8}))^2=1\)

\(\iff (\frac{\sqrt{2+\sqrt{2}}}{2})^2+(sin (\frac{\pi}{8}))^2=1\)

\(\iff \frac{2+\sqrt{2}}{4}+(sin (\frac{\pi}{8}))^2=1\)

\(\iff (sin (\frac{\pi}{8}))^2=1-\frac{2+\sqrt{2}}{4}\)

\(\iff (sin (\frac{\pi}{8}))^2=\frac{4}{4}-\frac{2+\sqrt{2}}{4}\)

\(\iff (sin (\frac{\pi}{8}))^2=\frac{4-(2+\sqrt{2})}{4}\)

\(\iff (sin (\frac{\pi}{8}))^2=\frac{4-2-\sqrt{2}}{4}\)

\(\iff (sin (\frac{\pi}{8}))^2=\frac{2-\sqrt{2})}{4}\)

\(\iff sin (\frac{\pi}{8})=\sqrt{\frac{2-\sqrt{2}}{4}}\) ou \(sin (\frac{\pi}{8})=-\sqrt{\frac{2-\sqrt{2}}{4}}\)

\(\iff sin (\frac{\pi}{8})=\frac{\sqrt{2-\sqrt{2}}}{2}\) ou \(sin (\frac{\pi}{8})=\frac{-\sqrt{2-\sqrt{2}}}{2}\)

or \(\frac{\pi}{8}\in[0 ; \frac{\pi}{2}]\) donc \(sin(\frac{\pi}{8})\ge 0\)

donc \(sin (\frac{\pi}{8})=\frac{\sqrt{2-\sqrt{2}}}{2}\)

2. \(cos(\frac{5\pi}{8})=cos(\frac{4\pi}{8}+\frac{\pi}{8})=cos(\frac{\pi}{2}+\frac{\pi}{8})=-sin(\frac{\pi}{8})\)

\(\iff cos(\frac{5\pi}{8})=-\frac{\sqrt{2-\sqrt{2}}}{2}\)

3.\(tan(\frac{\pi}{8})=\frac{sin(\frac{\pi}{8})}{cos(\frac{\pi}{8})}\)

\(\iff tan(\frac{\pi}{8})=\frac{\frac{\sqrt{2-\sqrt{2}}}{2}}{\frac{\sqrt{2+\sqrt{2}}}{2}}\)

3\(\iff tan(\frac{\pi}{8})=\frac{\sqrt{2-\sqrt{2}}}{2} \times \frac{2}{\sqrt{2+\sqrt{2}}}\)

\(\iff tan(\frac{\pi}{8})=\frac{\sqrt{2-\sqrt{2}}}{\sqrt{2+\sqrt{2}}}\)

\(\iff tan(\frac{\pi}{8})=\frac{(\sqrt{2-\sqrt{2}})(\sqrt{2-\sqrt{2}})}{(\sqrt{2+\sqrt{2}})(\sqrt{2-\sqrt{2}})}\)

\(\iff tan(\frac{\pi}{8})=\frac{2-\sqrt{2}}{\sqrt{2^2-\sqrt{2}^2}}\)

\(\iff tan(\frac{\pi}{8})=\frac{2-\sqrt{2}}{\sqrt{4-2}}\)

\(\iff tan(\frac{\pi}{8})=\frac{2-\sqrt{2}}{\sqrt{2}}\)

\(\iff tan(\frac{\pi}{8})=\frac{2}{\sqrt{2}}-\frac{\sqrt{2}}{\sqrt{2}}\)

\(\iff tan(\frac{\pi}{8})=\sqrt{2}-1\)

\((\sqrt{3-2\sqrt{2}})^2=3-2\sqrt{2}\)

\((\sqrt{2}-1)^2=3-2\sqrt{2}\)

donc \(\sqrt{3-2\sqrt{2}}=\sqrt{(\sqrt{2}-1)^2}=\sqrt{2}-1\)

Finalement \(tan(\frac{\pi}{8})=\sqrt{3-2\sqrt{2}}\)

\(2cos^2 x+2\sqrt{2} cos⁡ x+1=0\)

Posons \(X=cosx\)

donc \(X \in [0 ;1]\)

\(2X^2+2\sqrt{2}X+1=0\)

\(\Delta=b^2-4ac\)

\(\iff \Delta=(2\sqrt{2})^2-4\times 2 \times 1\)

\(\iff \Delta= 4\times 2-4\times 2\)

\(\iff \Delta=8-8\)

\(\iff \Delta=0\)

donc l'équation n'admet qu'une seule solution :

\(X_0=\frac{-b}{2a}=\frac{-2\sqrt{2}}{2\times 2}\)

\(\iff X_0=\frac{-2\sqrt{2}}{4}\)

\(\iff X_0=\frac{-\sqrt{2}}{2}\)

\(\iff cos x=\frac{-\sqrt{2}}{2}\)

\(\iff x= \frac{\pi}{4}+2k\pi\) ou \(x= -\frac{\pi}{4}+2k\pi\) \(k\in \mathbb{Z}\)

\(2 sin^2⁡ x+5sin x-3=0\)

Posons \(X=sinx\)

donc \(X \in [0 ;1]\)

\(2X^2+5X-3=0\)

\(\Delta=b^2-4ac\)

\(\iff \Delta=5^2-4\times 2 \times (-3)\)

\(\iff \Delta=25+24\)

\(\iff \Delta=49\)

\(\iff \Delta=49\)

donc l'équation admet deux solutions :

\(\begin{cases}X_1=\frac{-b-\sqrt{\Delta}}{2a}\\X_2=\frac{-b+\sqrt{\Delta}}{2a}\end{cases}\)

\(\iff \begin{cases}X_1=\frac{-5-\sqrt{49}}{2\times 2}\\X_2=\frac{-5+\sqrt{49}}{2\times 2}\end{cases}\)

\(\iff \begin{cases}X_1=\frac{-5-7}{4}\\X_2=\frac{-5+7}{4}\end{cases}\)

\(\iff \begin{cases}X_1=\frac{-12}{4}\\X_2=\frac{2}{4}\end{cases}\)

\(\iff \begin{cases}X_1=-3\\X_2=\frac{1}{2}\end{cases}\)

or \(X \in [0 ;1]\) donc \(X_1=-3\) est impossible

\(X_2=\frac{1}{2}\)

\(\iff sin x =\frac{1}{2}\)

\(\iff x= \frac{\pi}{6}+2k\pi\) ou \(x= \pi-\frac{\pi}{6}+2k\pi\) \(k \in\mathbb{Z}\)