Question

1)sin(-pi/12) =
2)cos(-pi/12)=
3)tg(-pi/12)=
4)ctg(-pi/12)=​

Answer 1

Ответ:

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

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

$3)\ \ tg(-\dfrac{\pi}{12}}=\dfrac{sin(-\dfrac{\pi}{12})}{cos(-\dfrac{\pi}{12})}=\dfrac{\sqrt2-\sqrt6}{\sqrt2+\sqrt6}=\dfrac{(\sqrt2-\sqrt6)^2}{2-6}=-\dfrac{8-2\sqrt{12}}{4}=\\\\\\=-\dfrac{8-2\cdot 2\sqrt3}{4}=-\dfrac{-4\cdot (\sqrt3-2)}{4}=\sqrt3-2$

$4)\ \ ctg(-\dfrac{\pi}{12}}=\dfrac{1}{tg(-\dfrac{\pi}{12})}=\dfrac{1}{\sqrt3-2}=\dfrac{\sqrt3+2}{3-4}=\dfrac{\sqrt3+2}{-1}=-2-\sqrt3$