Question

хэлп тригонометрия пожалуйста ​

Answer 1

Объяснение:

$\displaystyle\\cos\alpha =\frac{24}{25} \ \ \ \ 270^0 < \alpha < 360^0\ \ \ \ sin2\alpha =?\ \ \ \ cos2\alpha =?\ \ \ \ ctg2\alpha =?\\\\sin^2\alpha +cos^2\alpha =1\\\\sin^2\alpha =1-cos^2\alpha =1-(\frac{24}{25})^2=1-\frac{576}{625} =\frac{49}{625}. \\\\sin\alpha =б\sqrt{\frac{49}{625} } =б\frac{7}{25},\\\\270^0 < \alpha < 360^0\\\\sin\alpha =-\frac{7}{25}.$

$\displaystyle\\sin2\alpha =2*sin\alpha *cos\alpha =2*(-\frac{7}{25})*\frac{24}{25}=-\frac{336}{625}.\\\\ cos2\alpha =cos^2\alpha -sin^2\alpha =(\frac{24}{25} )^2-(-\frac{7}{25})^2=\frac{576}{625}-\frac{49}{625}=\frac{527}{625}.\\\\ ctg2\alpha =\frac{cos2\alpha }{sin2\alpha } =\frac{\frac{527}{625} }{\frac{-336}{625} } =-\frac{527}{336} .$