Question

80,БАЛЛОВ
Тригонометрическая функция

Answer 1

Ответ:

6

$\sin(360^{\circ} ) tg(160^{\circ} ) + \cos(180^{\circ} ) \sin(150^{\circ} ) + \sin(120^{\circ} ) \cos(150^{\circ} ) \\ \\ \sin(360^{\circ} ) = 0 \\ = > \sin(360^{\circ} ) tg(160^{\circ} ) = 0 \\ \\ \cos(180^{\circ} ) \sin(150^{\circ} ) + \cos(150^{\circ} ) \sin(120^{\circ} ) = \\ = - 1 \times \sin(180^{\circ} - 30^{\circ} ) + \cos(180^{\circ} - 30^{\circ} ) \sin(180^{\circ} - 60^{\circ} ) = \\ = - \sin(30^{\circ} ) - \cos(30^{\circ} ) \sin(60^{\circ} ) = - \frac{1}{2} - \frac{ \sqrt{3} }{2} \times \frac{ \sqrt{3} }{2} = \\ = - 0.5 - 1.5 = - 2$

4

$\frac{ \sin( \alpha ) }{ \cos(180^{\circ} - \alpha ) } - \frac{1}{tg(90^{\circ} - \alpha )} + 2 \sin( 90^{\circ} - \alpha ) = \\ = \frac{ \sin( \alpha ) }{ - \cos( \alpha ) } - \frac{1}{ctg( \alpha )} + 2 \cos( \alpha ) = \\ = - tg( \alpha ) + tg( \alpha ) + 2 \cos( \alpha ) = 2 \cos( \alpha )$