Question

Помогите плиз с матаном :^​

Answer 1

$x \: prinadlejit \: 4oy \: cetverti \\ \sin(x) = - \frac{5}{13} \\ {cos}^{2} x + {sin}^{2} x = 1 \\ {cos}x = \sqrt{1 - {sin }^{2} x} = \\ = \sqrt{1 - ( { \frac{5}{13} }^{2}) } = \sqrt{1 - \frac{25}{169} } = \\ = \sqrt{ \frac{144}{169} } = \frac{12}{13}$

$1)cos2x = { (\frac{12}{13}) }^{2} - { ( - \frac{5}{13}) }^{2} = \\ = \frac{144}{169} - \frac{25}{169} = \frac{119}{169}$

$tgx = \frac{sinx}{cosx} = \frac{ - \frac{5}{13} }{ \frac{12}{13} } = \\ = - \frac{5}{12}$

$2)tg2x = \frac{2tgx}{1 - {tg}^{2} x} = \\ = \frac{2( - \frac{5}{12} )}{1 - {( - \frac{5}{12}) }^{2} } = - \frac{ \frac{5}{6} }{ \frac{119}{144} } = \\ = - \frac{5 \times 144}{6 \times 119} = - \frac{120}{119}$

$ctgx = \frac{cosx}{sinx} = \\ = \frac{ \frac{12}{13} }{ - \frac{5}{13} } = - \frac{12}{5}$

$3)ctg2x = \frac{ {ctg}^{2} x - 1}{2ctgx} = \\ = \frac{ \frac{144}{25} - 1}{2 \times ( - \frac{12}{5}) } = \\ = - \frac{ \frac{119}{25} }{ \frac{24}{5} } = - \frac{119 \times 5}{25 \times 24} = - \frac{119}{120}$

4)sin2x=2sinxcosx=2×(-5/13)×(12/13)=

=-120/169

5)cos(п/6+х)=cos(п/6) cosх-sin(п/6) sinх=

=(корень3)/2 * (12/13)-(1/2)*(-5/13)=

=[12(корень3)+5]/26