Question

Знайди sin a і tg a якщо cos a =

Answer 1

Ответ:

1.

$\sin^{2}( \alpha ) + \cos^{2} ( \alpha ) = 1 \\ \sin^{2} ( \alpha ) = 1 - \cos^{2} ( \alpha ) \\ \sin( \alpha ) = \sqrt{1 - \cos^{2} ( \alpha ) }$

$\displaystyle \sin( \alpha ) = \sqrt{1 - {\bigg( - \frac{1}{2} \bigg)}^{2} } \\ \sin( \alpha ) = \sqrt{1 - \frac{1}{4} } \\ \sin( \alpha ) = \sqrt{ \frac{3}{4} } \\ \sin( \alpha ) = \frac{ \sqrt{3} }{2}$

$\displaystyle \tan( \alpha ) = \frac{ \sin( \alpha ) }{ \cos( \alpha ) }$

$\displaystyle \tan( \alpha ) = \frac{ \frac{ \sqrt{3} }{2} }{ - \frac{1}{2} } = \frac{ \sqrt{3} }{2} \times ( - 2) = - \sqrt{3}$

2.

$\sin( \beta ) > \sin( \alpha ) \: \: \: = >$

$= > \: \: \: \beta > \alpha$

3. а)

$\tan( \alpha ) \cos( \alpha ) + \sin( \alpha) = 2 \sin( \alpha ) \\ \frac{ \sin( \alpha ) }{ \cos( \alpha ) } \times \cos( \alpha ) + \sin( \alpha ) = 2 \sin( \alpha ) \\ \sin( \alpha ) + \sin( \alpha ) = 2 \sin( \alpha ) \\ 2 \sin( \alpha ) = 2 \sin( \alpha )$

б)

$(1 - \sin ^{2} ( \alpha ) ) \tan^{2} ( \alpha ) = \sin^{2} ( \alpha ) \\ \cos ^{2} ( \alpha ) \times \frac{ \sin^{2} ( \alpha ) }{ \cos^{2} ( \alpha ) } = \sin^{2} ( \alpha ) \\ \sin ^{2} ( \alpha ) = \sin^{2} ( \alpha )$

в)

$(1 - \cos( \alpha )) (1 + \cos( \alpha )) = \sin^{2} ( \alpha ) \\ 1 - \cos ^{2} ( \alpha ) = \sin^{2} ( \alpha ) \\ \sin^{2} ( \alpha ) = \sin ^{2} ( \alpha )$

г)

$\frac{ \cos( {90}^{ \circ} - \alpha ) }{ \cos( {180}^{ \circ} - \alpha ) } = - \tan( \alpha ) \\ \frac{ \sin( \alpha ) }{ - \cos( \alpha ) } = - \tan( \alpha ) \\ - \tan( \alpha ) = - \tan( \alpha )$