Réponse:
On utilise cos²x + sin²x = 1
[tex] {cos}^{2} ( \frac{7\pi}{12} ) = 1 - {sin}^{2} ( \frac{7\pi}{12} ) [/tex]
[tex]{cos}^{2} ( \frac{7\pi}{12} ) = 1 - \frac{ {( \sqrt{2} + \sqrt{6}) }^{2} }{ {4}^{2} } [/tex]
[tex]{cos}^{2} ( \frac{7\pi}{12} ) = 1 - \frac{ {(2 + 2 \times \sqrt{2} \times \sqrt{6} + 6) }}{ 16 } [/tex]
[tex]{cos}^{2} ( \frac{7\pi}{12} ) = \frac{ {(16 - 8 - 2 \times \sqrt{2} \times \sqrt{6}) }}{16 } [/tex]
[tex]{cos}^{2} ( \frac{7\pi}{12} ) = \frac{ {(8 - 2 \times \sqrt{2} \times \sqrt{6}) }}{16 } [/tex]
[tex]{cos}^{2} ( \frac{7\pi}{12} ) = \frac{ {(2 + 6 - 2 \times \sqrt{2} \times \sqrt{6}) }}{16 } [/tex]
[tex]{cos}^{2} ( \frac{7\pi}{12} ) = \frac{ {( \sqrt{2} - \sqrt{6}) }^{2} }{ {4}^{2} } [/tex]
2.
[tex] \frac{7\pi}{12} \geqslant \frac{\pi}{2} [/tex]
donc son cosinus est negatif.
[tex]{cos}( \frac{7\pi}{12} ) = - \sqrt{ \frac{ {( \sqrt{2} - \sqrt{6}) ^{2} }}{ 4^{2}} }[/tex]
[tex]{cos}( \frac{7\pi}{12} ) = - \frac{ {( \sqrt{2} - \sqrt{6}) }}{ 4} [/tex]
[tex]{cos}( \frac{7\pi}{12} ) = \frac{ {\sqrt{6} - \sqrt{2}}}{ 4} [/tex]
