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

Answer 2

отметила же тут))

Answer 3

это полностью решенное уравнение?