Question

вычислите. sin2α, cos2α, tg2α ,если: а)sinα=-3/5 и π <α<3π /2. б)cosα=5/13 и 3π /2 <α<2π . в)tgα=-3/4 и π/2<α<π .

Answer 1

а)
sina = -3/5
$cosa = - sqrt{1-sin^2a} = -sqrt{1-9/25} = -sqrt{16/25} = -4/5$
$sin2a = 2sina*cosa = 2*frac{3}{5}*frac{4}{5} = frac{24}{25}$
$cos2a = cos^2a - sin^2a = frac{16}{25}-frac{9}{25} = frac{7}{25}$
tga = sina/cosa = 3/5 : 4/5 = 3/4
$tg2a = frac{2tga}{1-tg^2a} = frac{2*3}{4}:(1-frac{9}{16}) =frac{3}{2}:frac{7}{16} = frac{3*16}{2*7} = frac{24}{7}$ 

б)
cosa = 5/13
$sina = - sqrt{1-cos^2a} = -sqrt{1-25/169} = -sqrt{144/169} = -12/13$
$sin2a = 2sina*cosa = - 2*frac{5}{13}*frac{12}{13} = frac{24}{25}$

$cos2a = cos^2a - sin^2a = frac{25}{169} - frac{144}{169} = -frac{119}{169}$ 

tga = sina/cosa = -12/13 : 5/13 = -12/5

$tg2a = frac{2tga}{1-tg^2a} = -frac{24}{5} : (1-frac{144}{25}) = frac{24*25}{5*119} =frac{120}{119}$ 

 

в)

$sin2a = frac{2tga}{1+tg^2a} = -frac{3}{2}:(1+frac{9}{16}) = -frac{3}{2}:frac{25}{16} = -frac{3*8}{25} = -frac{24}{25}$ 

$cos2a = frac{1-tg^2a}{1+tg^2a} = (1-frac{9}{16}):(1+frac{9}{16}) = frac{7}{16}:frac{25}{16} = frac{7}{25}$ 

$tg2a = frac{2tga}{1-tg^2a} = -frac{3}{2}(1-frac{9}{16}) = -frac{3}{2}:frac{7}{16} = -frac{24}{7}$