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\begin{array}{l} Sans calculer cos \frac{π}{8} et sin \frac{π}{8}, démontrer que : \\ \frac{cos \frac{π}{8} + sin \frac{π}{8}}{cos \frac{π}{8} - sin \frac{π}{8}} = 1 + \sqrt{2} \end{array}

\begin{array}{l} \frac{cos \frac{π}{8} + sin \frac{π}{8}}{cos \frac{π}{8} - sin \frac{π}{8}} = \frac{(cos \frac{π}{8} + sin \frac{π}{8}) (cos \frac{π}{8} + sin \frac{π}{8})}{(cos \frac{π}{8} - sin \frac{π}{8}) (cos \frac{π}{8} + sin \frac{π}{8})} \\ = \frac{{(cos \frac{π}{8} + sin \frac{π}{8})}^{2}}{{cos}^{2} \frac{π}{8} - {sin}^{2} \frac{π}{8}} = \frac{{cos}^{2} \frac{π}{8} + {sin}^{2} \frac{π}{8} + 2sin \frac{π}{8} cos \frac{π}{8}}{cos (2 \times \frac{π}{8})} \\ \{car cos 2a = {cos}^{2} a - {sin}^{2} a\} \\ = \frac{1 + sin \frac{2 \times π}{8}}{cos \frac{π}{4}} \{car {cos}^{2} a + {sin}^{2} a = 1     et   sin 2a = 2 sin a cos a\} \\ = \frac{1 + sin \frac{π}{4}}{cos \frac{π}{4}} = \frac{1 + \frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = \frac{2 \sqrt{2} + 2}{2} = 1 + \sqrt{2} \end{array}

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