Question

вычислить cos a, tg a, ctg a, если sin a -5/13,
3п/2<а<2п

Answer 1

Объяснение:

$sin\alpha =-\frac{5}{13}\ \ \ \ \frac{3\pi }{2} <\alpha <2\pi \ \ \ \ cos\alpha =?\ \ \ tg\alpha =?\ \ \ ctg\alpha =?\\sin^2\alpha +cos^2\alpha =1\\cos^2=1-sin^2\alpha =1-(-\frac{5}{13})^2=1- \frac{25}{169} =\frac{169-25}{169}=\frac{144}{169}.$

$cos\alpha =б\sqrt{\frac{144}{169} }=б\frac{12}{13}\\cos\alpha= \frac{12}{13}\ \ (\frac{3\pi }{2} <\alpha <2\pi) \\tg\alpha =\frac{sin\alpha }{cos\alpha } =\frac{-\frac{5}{13} }{\frac{12}{13} } =-\frac{5}{13}.\\ctg\alpha =\frac{1}{tg\alpha } =-\frac{13}{5} .$