Question

2sin п/12-сos п/12срочноооо решитеееее пожалуйста через 10мин

Answer 1

$\frac{\pi }{12}=\frac{180^\circ }{12}=15^\circ \\\\\\sin15^\circ =sin(45^\circ -30^\circ )=sin45^\circ\cdot cos30^\circ -sin30^\circ \cdot cos45^\circ =\\\\=\frac{\sqrt2}{2}\cdot \frac{\sqrt3}{2}-\frac{1}{2}\cdot \frac{\sqrt2}{2}=\frac{\sqrt6-\sqrt2}{4}=\frac{\sqrt2(\sqrt3-1)}{4}\\\\\\cos15^\circ =cos(45^\circ -30^\circ )=cos45^\circ \cdot cos30^\circ +sin45^\circ \cdot sin30^\circ =\\\\=\frac{\sqrt2}{2}\cdot \frac{\sqrt3}{2}+\frac{\sqrt2}{2}\cdot \frac{1}{2}=\frac{\sqrt6+\sqrt2}{4}=\frac{\sqrt2(\sqrt3+1)}{4}$

$2\, sin\frac{\pi }{12}-cos\frac{\pi }{12}=2\cdot \frac{\sqrt6-\sqrt2}{4}-\frac{\sqrt6+\sqrt2}{4}=\frac{2\sqrt6-2\sqrt2-\sqrt6-\sqrt2}{4}=\\\\=\frac{\sqrt6-3\sqrt2}{4}=\frac{\sqrt2(\sqrt3-3)}{4}$