Question

Самостоятельная по тригонометрии, помогите пж, 35 баллов

Answer 1

Ответ:

1.

$\sin( \alpha ) = - 0.8$

угол принадлежит 4 четверти, косинус > 0

$\cos( \alpha ) = \sqrt{1 - \sin {}^{2} ( \alpha ) } \\ \cos( \alpha ) = \sqrt{1 - 0.64} = \sqrt{0.36} = 0.6$

$tg \alpha = \frac{ \sin( \alpha ) }{ \cos( \alpha ) } = \frac{ - 0.8}{0.6} = - \frac{8}{10} \times \frac{10}{6} = - \frac{4}{3}$

$ctg \alpha = \frac{1}{tg \alpha } = - \frac{3}{4}$

2.

${(1 + tg \beta )}^{2} - 2tg \beta = 1 + 2tg \beta +t g {}^{2} \beta - 2tg \beta = \\ = 1 + {tg}^{2} \beta = \frac{1}{ \cos {}^{2} ( \beta ) }$

3.

$\frac{ \cos(0) \sin {}^{2} (\phi) }{(tg \frac{\pi}{4} - \sin(\phi)) ( \sin( \frac{\pi}{2}) + \sin(\phi)) ) } = \\ = \frac{ \sin {}^{2} (\phi) }{(1 - \sin(\phi)) (1 + \sin(\phi)) } = \frac{ \sin {}^{2} (\phi) }{1 - { \sin }^{2}(\phi) } = \\ = \frac{ \sin {}^{2} (\phi) }{ \cos {}^{2} (\phi) } = {tg}^{2}( \phi)$

4.

$\frac{ctg \gamma }{ \cos( \gamma ) } - \frac{ \cos( \gamma ) }{tg \gamma } = \frac{ \cos( \gamma ) }{ \sin( \gamma ) } \times \frac{1}{ \cos( \gamma ) } - \cos( \gamma ) \times \frac{ \cos( \gamma ) }{ \sin( \gamma ) } = \\ = \frac{1}{ \sin( \gamma ) } - \frac{ \cos {}^{2} ( \gamma ) }{ \sin( \gamma ) } = \frac{ \sin {}^{2} ( \gamma ) }{ \sin {}^{} ( \gamma ) } = \sin( \gamma )$