Question

ПОМОГИТЕ ПРОШУ! ДАЮ 35 баллов!

Answer 1

Ответ:

$1)\ \ cos\dfrac{\pi}{10}\cdot cos\dfrac{2\pi}{5}-sin\dfrac{\pi}{10}\cdot sin\dfrac{2\pi}{5}=cos\Big(\dfrac{\pi}{10}+\dfrac{2\pi}{5}\Big)=cos\dfrac{5\pi}{10}=cos\dfrac{\pi}{2}=0\\\\\\2)\ \ sin\dfrac{\pi}{6}\cdot cos\dfrac{\pi}{12}+cos\dfrac{\pi}{6}+sin\dfrac{\pi}{12}=\dfrac{1}{2}\cdot\dfrac{\sqrt6+\sqrt2}{4}+\dfrac{\sqrt3}{2}+\dfrac{\sqrt6-\sqrt2}{4}=\\\\\\=\dfrac{\sqrt6+\sqrt2+4\sqrt3+2\sqrt6-2\sqrt2}{8}=\dfrac{3\sqrt6-\sqrt2+4\sqrt3}{8}$

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

$3)\ \ cos50^\circ \cdot cos5^\circ +sin50^\circ \cdot sin5^\circ =cos(50^\circ -5^\circ )=cos45^\circ =\dfrac{\sqrt2}{2}\\\\\\4)\ \ sin71^\circ \cdot cos11^\circ -cos71^\circ \cdot \cdot sin11^\circ =sin(71^\circ -11^\circ )=sin60^\circ =\dfrac{\sqrt3}{2}$