Question

ДОПОМОЖІТЬ с алгеброю будь-ласка

Answer 1

Объяснение:

$\displaystyle\\cos\alpha =\frac{4}{5} \ \ \ \ \frac{3\pi }{2} < \alpha < 2\pi \ \ \ \ sin\alpha =?\ \ \ \ tg\alpha =?\ \ \ \ ctg\alpha =?\ \ \ \ sin2\alpha =?\ \ \ \ cos2\alpha =?\\\\sin^2\alpha +cos^2\alpha =1\\\\sin^2\alpha =1-cos^2\alpha =1-(\frac{4}{5})^2=1-\frac{16}{25} =\frac{25*1-16}{25} =\frac{9}{25} .\\\\sin\alpha =б\sqrt{\frac{9}{25} } =б\frac{3}{5}.\\\\\frac{3\pi }{2} < \alpha < 2\pi \ \ \ \ \ \ \Rightarrow\\\\sin\alpha =-\frac{3}{5} .$$\displaystyle\\tg\alpha =\frac{sin\alpha }{cos\alpha } =\frac{-\frac{3}{5} }{\frac{4}{5} } =-\frac{3}{4}.\\\\ctg\alpha =\frac{1}{tg\alpha }=-\frac{4}{3} .$

$\displaystyle\\sin2\alpha =2*sin\alpha* cos\alpha =2*(-\frac{3}{5})*\frac{4}{5}=-\frac{24}{25} .\\\\cos2\alpha =cos^2\alpha -sin^2\alpha =(\frac{4}{5})^2-(-\frac{3}{5} )^2=\frac{16}{25} -\frac{9}{25}=\frac{16-9}{24}=\frac{7}{25} .$