Question

значения тригонометрических функций

Answer 1

$1) \: \: \sin( \alpha ) = \sqrt{1 - { \cos( \alpha ) }^{2} }$
$\sin( \alpha ) = \sqrt{ \frac{169}{169} - \frac{144}{169} } = \sqrt{ \frac{25}{169} }$
Sin a = 5/13
$\tan( \alpha ) = \frac{ \sin( \alpha ) }{ \cos( \alpha ) }$
Tg a = (5/13)/(12/13)
tg a = (5×13)/(13×12)=5/12

$\cot( \alpha ) = \frac{1}{ \tan( \alpha ) }$
Ctg a = 12/5

2)
$\cos( \alpha ) = - \sqrt{1 - { \sin( \alpha ) }^{2} }$
$\cos( \alpha ) = - \sqrt{ \frac{16}{16} - \frac{3}{16} } = - \sqrt{ \frac{13}{16} }$

$\cos( \alpha ) = - \frac{ \sqrt{13} }{4}$
$\tan( \alpha ) = \frac{ \sqrt{3} }{4} \div \frac{ \sqrt{13} }{4} = \frac{ \sqrt{3} \times 4 }{ \sqrt{13} \times 4}$
$\tan( \alpha ) = \frac{ \sqrt{3} }{ \sqrt{13} }$
$\cot( \alpha ) = \frac{ \sqrt{13} }{ \sqrt{3} }$
3) ctg a= - 3

$1 + { \cot( \alpha ) }^{2} = \frac{1}{ { \sin( \alpha ) }^{2} }$
${ \sin( \alpha ) }^{2} = \frac{1}{1 + { \cot( \alpha ) }^{2} }$
$\sin( \alpha ) = - \frac{1}{ \sqrt{1 + { \cot( \alpha ) }^{2} } }$
$\sin( \alpha ) = - \frac{1}{ \sqrt{1 + 9} } = - \frac{1}{ \sqrt{10} }$
$\cos( \alpha ) = \sqrt{1 - { \sin( \alpha ) }^{2} }$
$\cos( \alpha ) = \sqrt{ \frac{10}{10} - \frac{1}{10} } = \sqrt{ \frac{9}{10} }$
$\cos( \alpha ) = \frac{3}{ \sqrt{10} }$
4) tg a=-1/7
$\sin( \alpha ) = \frac{1}{ \sqrt{1 + 49} } = \frac{1}{ \sqrt{50} } = \frac{1}{5 \sqrt{2} }$
$\cos( \alpha ) = - \sqrt{ \frac{50}{50} - \frac{1}{50} } = - \sqrt{ \frac{49}{50} }$
$\cos( \alpha ) = - \frac{7}{5 \sqrt{2} }$