Exercice : Exercice 21 :
Question
1.En remarquant l'égalité
\(\frac{\pi}{12}=\frac{\pi}{3}-\frac{\pi}{4}\)
déterminer les valeurs de \(cos(\frac{\pi}{12})\) et \(sin(\frac{\pi}{12})\)
Indice
\(cos(a-b)=cos(a)cos(b)+sin(a)sin(b)\)
\(sin(a-b)=sin(a)cos(b)-cos(a)sin(b)\)
Solution
\(\frac{\pi}{3}-\frac{\pi}{4}=\frac{4\pi}{12}-\frac{3\pi}{12}=\frac{\pi}{12}\)
\(cos(\frac{\pi}{12})=cos(\frac{\pi}{3}-\frac{\pi}{4})=cos(\frac{\pi}{3})cos(\frac{\pi}{4})+sin(\frac{\pi}{3})sin(\frac{\pi}{4})=\frac{1}{2} \times \frac{\sqrt{2}}{2}+\frac{\sqrt{3}}{2} \times \frac{\sqrt{2}}{2}\)
\(cos(\frac{\pi}{12})=\frac{\sqrt{2}}{4}+\frac{\sqrt{3} \times \sqrt{2}}{4}=\frac{\sqrt{2}(1+\sqrt{3})}{4}\)
\(\color{red}{cos(\frac{\pi}{12})=\frac{\sqrt{2}(1+\sqrt{3})}{4}}\)
\(\color{magenta}{\textbf{Rappel à l'exercice 22, on a trouvé :}}\)
\(\color{red}{cos(\frac{\pi}{12})=\frac{\sqrt{2+\sqrt{3}}}{2}}\)
Montrons que les deux valeurs sont égales :
\(cos(\frac{\pi}{12})=\frac{\sqrt{4}\sqrt{2+\sqrt{3}}}{4}\)
\(cos(\frac{\pi}{12})=\frac{\sqrt{2}\times \sqrt{2(2+\sqrt{3})}}{4}\)
\(cos(\frac{\pi}{12})=\frac{\sqrt{2}\times \sqrt{4+2\sqrt{3}}}{4}\)
\(cos(\frac{\pi}{12})=\frac{\sqrt{2}\times \sqrt{1+2\sqrt{3}+3}}{4}\)
\(cos(\frac{\pi}{12})=\frac{\sqrt{2}\times \sqrt{1+2\sqrt{3}+3}}{4}\)
\(cos(\frac{\pi}{12})=\frac{\sqrt{2}\times \sqrt{(1+\sqrt{3})^2}}{4}\)
\(cos(\frac{\pi}{12})=\frac{\sqrt{2}\times (1+\sqrt{3})}{4}\)
Les deux valeurs sont donc égales.
\(\frac{\pi}{3}-\frac{\pi}{4}=\frac{4\pi}{12}-\frac{3\pi}{12}=\frac{\pi}{12}\)
\(sin(\frac{\pi}{12})=sin(\frac{\pi}{3}-\frac{\pi}{4})=sin(\frac{\pi}{3})cos(\frac{\pi}{4})-cos(\frac{\pi}{3})sin(\frac{\pi}{4})=\frac{\sqrt{3}}{2} \times \frac{\sqrt{2}}{2}-\frac{1}{2} \times \frac{\sqrt{2}}{2}\)
\(cos(\frac{\pi}{12})=\frac{\sqrt{3} \times \sqrt{2}}{4} - \frac{\sqrt{2}}{4}=\frac{\sqrt{2}(\sqrt{3}-1)}{4}\)
\(\color{red}{cos(\frac{\pi}{12})=\frac{\sqrt{2}(\sqrt{3}-1)}{4}}\)
Question
2.Déterminer les valeurs de
\(cos(\frac{7\pi}{12})\) et \(sin(\frac{7\pi}{12})\)
Indice
\(cos(a+b)=cos(a)cos(b)-sin(a)sin(b)\)
\(sin(a+b)=sin(a)cos(b)+cos(a)sin(b)\)
Solution
\(\frac{\pi}{3}+\frac{\pi}{4}=\frac{4\pi}{12}+\frac{3\pi}{12}=\frac{7\pi}{12}\)
\(cos(\frac{7\pi}{12})=cos(\frac{\pi}{3}+\frac{\pi}{4})=cos(\frac{\pi}{3})cos(\frac{\pi}{4})-sin(\frac{\pi}{3})sin(\frac{\pi}{4})=\frac{1}{2} \times \frac{\sqrt{2}}{2}-\frac{\sqrt{3}}{2} \times \frac{\sqrt{2}}{2}\)
\(cos(\frac{7\pi}{12})=\frac{\sqrt{2}}{4}-\frac{\sqrt{3} \times \sqrt{2}}{4}=\frac{\sqrt{2}(1-\sqrt{3})}{4}\)
\(\color{red}{cos(\frac{7\pi}{12})=\frac{\sqrt{2}(1-\sqrt{3})}{4}}\)
\(\frac{\pi}{3}+\frac{\pi}{4}=\frac{4\pi}{12}+\frac{3\pi}{12}=\frac{7\pi}{12}\)
\(sin(\frac{7\pi}{12})=sin(\frac{\pi}{3}+\frac{\pi}{4})=sin(\frac{\pi}{3})cos(\frac{\pi}{4})+cos(\frac{\pi}{3})sin(\frac{\pi}{4})=\frac{\sqrt{3}}{2} \times \frac{\sqrt{2}}{2}+\frac{1}{2} \times \frac{\sqrt{2}}{2}\)
\(sin(\frac{7\pi}{12})=\frac{\sqrt{3} \times \sqrt{2}}{4} + \frac{\sqrt{2}}{4}=\frac{\sqrt{2}(\sqrt{3}+1)}{4}\)
\(\color{red}{sin(\frac{7\pi}{12})=\frac{\sqrt{2}(\sqrt{3}+1)}{4}}\)