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 .

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Ответы и объяснения

2012-03-10T15:39:39+00:00

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}