Question

Дано:
cos t= -5/13
pi/2 < t < pi
Найти:
sin t,tg t и ctg t

Answer 1

$sin alpha = sqrt{1-cos^2 alpha } \ \ sint= sqrt{1-frac{25}{169} } = sqrt{ frac{169-25}{169} } = sqrt{ frac{144}{169} }=frac{12}{13} \ \ tg alpha = frac{sin alpha }{cos alpha } \ \ tgt= frac{frac{12}{13}}{frac{-5}{13} } =frac{12}{13}*frac{13}{-5}=frac{12}{-5}=-2,4 \ \ ctg alpha = frac{cos alpha }{sin alpha } \ \ ctgt=frac{frac{-5}{13}}{frac{12}{13}}=frac{-5}{13}*frac{13}{12}=frac{-5}{12}=-frac{5}{12}$