Question

АЛГЕБРА 10 класс, 27 БАЛЛОВ

Answer 1

Объяснение:

$36.\ \frac{10*sin40^0*sin50^0}{cos10^0}=10*\frac{\frac{1}{2}*(cos(50^0-40^0)-cos(50^0+40^0) }{cos10^0} =5*\frac{cos10^0-cos90^0}{cos10^0}=\\=5*\frac{cos10^0-0}{cos10^0}=5*\frac{cos10^0}{cos10^0}=5.\\38.\ sin^2\alpha +cos(60^0+\alpha )*cos(60^0-\alpha )=\\ =sin^2\alpha +\frac{1}{2}*(cos(60^0+\alpha -60^0+\alpha )+ cos(60^0+\alpha-60^0-\alpha )=$

$=sin^2\alpha +\frac{1}{2}*(cos2\alpha +cos120^0)=\frac{2*sin^2\alpha +cos2\alpha +(-\frac{1}{2} )}{2}= \frac{2*sin^2\alpha +cos^2\alpha -sin^2\alpha -\frac{1}{2} }{2}=\\=\frac{sin^2\alpha +cos^2\alpha -\frac{1}{2} }{2} =\frac{1-\frac{1}{2} }{2}=\frac{\frac{1}{2} }{2}=\frac{1}{4}.$

$39.\ \frac{2*sin^2\alpha -1}{1-2*cos^2\alpha } =\frac{2*sin^2\alpha -sin^2\alpha -cos^2\alpha }{sin^2\alpha +cos^2\alpha -2*cos^2\alpha } =\frac{sin^2\alpha -cos^2\alpha }{sin^2\alpha -cos^2\alpha } =1.$