Question

Алгебра, 10.

Заполнить таблицу:

Answer 1

sina

$\sin(720°) = \sin(360° \times 2) = 0$

$\sin(225°) = \sin(180° + 45°) = - \sin(45°) = - \frac{ \sqrt{2} }{2}$

$\sin(300°) = \sin(270° + 30°) = - \cos(30°) = - \frac{ \sqrt{3} }{2}$

$\sin(870°) = \sin(720° + 150°) = \sin(150°) = \\ = \sin(180 °- 30°) = \sin(30°) = \frac{1}{2}$

$\sin(900°) = \sin(720° + 180°) = \sin(180°) = 0$

$\sin( - 330°) = \sin( - 360 °+ 30°) = \sin(30°) = \frac{1}{2}$

$\sin( - 630°) = \sin( - 720 °+ 90°) = \sin(90) = 1$

$\sin( - 210°) = \sin( - 180° - 30°) = \sin(30°) = \frac{1}{2}$

cosa

$\cos(720°) = 1$

$\cos(225°) = \cos(180° + 45°) = - \cos(45°) = - \frac{ \sqrt{2} }{2}$

$\cos(300°) = \cos(270 °+ 30°) = \sin(30°) = \frac{ 1}{2}$

$\cos(870°) = \cos(720 °+ 150°) = \cos(150°) = \\ = \cos(180° - 30°) = - \cos(30°) = - \frac{ \sqrt{3} }{2}$

$\cos(900°) = \cos(180°) = - 1 \\ \cos( - 330°) = \cos( - 360° + 30°) = \\ = \cos(30°) = \frac{ \sqrt{3} }{2}$

$\cos( - 630°) = \cos( - 720 °+ 90°) = \cos(90°) = 0$

$\cos( - 210°) = \cos( - 180° - 30°) = - \cos(30°) = - \frac{ \sqrt{3} }{2}$

tga

$tg \alpha = \frac{ \sin( \alpha ) }{ \cos( \alpha ) }$

$tg(720°) = 0$

$tg(225°) = - \frac{ \sqrt{2} }{2} \times ( - \frac{2}{ \sqrt{2} } ) = 1$

$tg(300°) = - \frac{ \sqrt{3} }{2} \times \frac{2}{ 1} = - \sqrt{3}$

$tg(870°) = \frac{1}{2} \times ( - \frac{2}{ \sqrt{3} } ) = \frac{1}{ \sqrt{3} } = - \frac{ \sqrt{3} }{3}$

$tg(900°) = 0 \times - 1 = 0 \\ tg( - 330°) = \frac{1}{2} \times \frac{2}{ \sqrt{3} } = \frac{1}{ \sqrt{3} } = \frac{ \sqrt{3} }{3}$

$tg( - 630°)$

не определен

$tg( - 210°) = \frac{1}{2} \times ( - \frac{2}{ \sqrt{3} } ) = - \frac{ \sqrt{3} }{3}$

ctga

$ctg \alpha = \frac{1}{tg \alpha }$

$ctg(720°)$

не определен

$ctg(225°) = 1 \\ ctg(300°) = - \frac{ \sqrt{3} }{3} \\ ctg(870°) = - \sqrt{3}$

$ctg(900°)$

не определен

$ctg( - 330°) = \sqrt{3} \\ ctg( - 630°) = 0 \\ ctg( - 210°) = - \sqrt{3}$