Exercice : Somme 3
Question
1.Après avoir vérifié que
\(\frac{7\pi}{12}=\frac{\pi}{4}+\frac{\pi}{3}\),
calculer \(cos\left(\frac{7\pi}{12}\right).\)
Indice
\(cos(a+b)=cos(a)cos(b)-sin(a)sin(b)\)
Solution
\(\frac{\pi}{4}+\frac{\pi}{3}=\frac{3\pi}{12}+\frac{4\pi}{12}=\frac{7\pi}{12}\)
donc
\(cos\left(\frac{7\pi}{12}\right)\)
\(=cos\left(\frac{\pi}{4}+\frac{\pi}{3}\right)\)
\(=cos\left(\frac{\pi}{4}+\frac{\pi}{3}\right)\)
\(=cos\left(\frac{\pi}{4}\right)cos\left(\frac{\pi}{3}\right)-sin\left(\frac{\pi}{4}\right)sin\left(\frac{\pi}{3}\right)\)
\(=\frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2} \times \frac{1}{2}\)
\(=\frac{\sqrt{6}}{4} -\frac{\sqrt{2}}{4}\)
\(=\frac{\sqrt{6}-\sqrt{2}}{4}\)
Question
2. En remarquant que \(\frac{\pi}{12}=\frac{\pi}{3} - \frac{\pi}{4}\)
Calculer \(sin(\frac{\pi}{12})\) et \(cos(\frac{\pi}{12})\)
Vérifier que \(sin^2(\frac{\pi}{12})+cos^2(\frac{\pi}{12})=1\)
Indice
\(sin(a-b)=sin(a)cos(b)-cos(a)sin(b)\)
\(cos(a-b)=cos(a)cos(b)+sin(a)sin(b)\)
Solution
\(\frac{\pi}{3} - \frac{\pi}{4}=\frac{4\pi}{12} - \frac{3\pi}{12}=\frac{\pi}{12}\)
\(sin(\frac{\pi}{12})=sin(\frac{\pi}{3} - \frac{\pi}{4})\)
\(=sin(\frac{\pi}{3})cos(\frac{\pi}{4})-cos(\frac{\pi}{3} )sin(\frac{\pi}{4})\)
\(=\frac{\sqrt{3}}{2} \times \frac{\sqrt{2}}{2}-\frac{1}{2} \times \frac{\sqrt{2}}{2}\)
\(=\frac{\sqrt{6}}{4}-\frac{\sqrt{2}}{4}\)
\(=\frac{\sqrt{6}-\sqrt{2}}{4}\)
\(cos(\frac{\pi}{12})=cos(\frac{\pi}{3} - \frac{\pi}{4})\)
\(=cos(\frac{\pi}{3})cos(\frac{\pi}{4})+sin(\frac{\pi}{3} )sin(\frac{\pi}{4})\)
\(=\frac{1}{2} \times \frac{\sqrt{2}}{2}+\frac{\sqrt{3}}{2} \times \frac{\sqrt{2}}{2}\)
\(=\frac{\sqrt{2}}{4}+\frac{\sqrt{6}}{4}\)
\(=\frac{\sqrt{2}+\sqrt{6}}{4}\)
\(sin^2(\frac{\pi}{12})+cos^2(\frac{\pi}{12})\)
\(=(\frac{\sqrt{6}-\sqrt{2}}{4})^2+(\frac{\sqrt{2}+\sqrt{6}}{4})^2\)
\(=\frac{(\sqrt{6}-\sqrt{2})^2}{16}+\frac{(\sqrt{2}+\sqrt{6})^2}{16}\)
\(=\frac{\sqrt{6}^2-2\sqrt{6} \times \sqrt{2}+ \sqrt{2}^2}{16}+\frac{\sqrt{2}^2+2\sqrt{2} \times \sqrt{6}+ \sqrt{6}^2}{16}\)
\(=\frac{6-2\sqrt{12}+ 2}{16}+\frac{2+2\sqrt{12}+6}{16}\)
\(=\frac{6-2\sqrt{12}+ 2+2+2\sqrt{12}+6}{16}\)
\(=\frac{6+ 2+2+6}{16}\)
\(=1\)
donc \(sin^2(\frac{\pi}{12})+cos^2(\frac{\pi}{12})=1\)