Question

Lucrez dintr-o culegere pentru admitere la facultate. Ajutati-ma cu TG 8.

Dau 50+25 puncte.

Answer 1

[tex]\displaystyle Se~arata~usor~ca~cos \frac{\pi}{12} =cos( \frac{\pi}{3} -\frac{\pi}{4})= \frac{\sqrt{6}+ \sqrt{2} }{4} \\ cos \frac{5\pi}{12} =cos( \frac{\pi}{2} -\frac{\pi}{12})=sin \frac{\pi}{12}= sin( \frac{\pi}{3} -\frac{\pi}{4})=\frac{\sqrt{6}- \sqrt{2} }{4} \\ A=cos^8(\frac{5\pi}{12})-cos^8(\frac{\pi}{12})=\\ =(\frac{\sqrt{6}- \sqrt{2} }{4} )^8-(\frac{\sqrt{6}+ \sqrt{2} }{4} )^8=\\ = \frac{(\sqrt{6}- \sqrt{2})^8-(\sqrt{6}+\sqrt{2})^8}{2^{16}} =\\ =\frac{4^4[(2- \sqrt{3})^4-(2+\sqrt{3})^4]}{2^{16}} [/tex]
$A=- \frac{ 7\sqrt{3} }{16}$