Question

вычислите значение тригонометрической функции

Answer 1

$ctg(2arccos dfrac{12}{13}- dfrac{5 pi }{2})$

Пусть $arccos dfrac{12}{13}=x$

$ctg(2x- dfrac{5 pi }{2})=-tg2x= dfrac{2tgx}{tg^2x-1} \ \ cosx= dfrac{12}{13} Rightarrow tgx= sqrt{ dfrac{1}{cos^2x}-1 }= sqrt{ dfrac{1}{( dfrac{12}{13} )^2}-1 } = sqrt{ dfrac{169}{144}-1 }= dfrac{5}{12} \ \ dfrac{2tgx}{tg^2x-1}= dfrac{2cdot dfrac{5}{12} }{( dfrac{5}{12})^2-1 }= dfrac{ dfrac{5}{6} }{- dfrac{119}{144} }=- dfrac{120}{119} - OTBET$