Question

Найдите значение остальных тригонометрических функций, если tgx =-5/12 и П/2

Answer 1

$tgx=-\frac5{12},\;\;\;\frac\pi2\leq x\leq\pi\\1)\;ctgx=\frac1{tgx}=\frac1{-\frac5{12}}=-\frac{12}5\\\\2)\;1+tg^2x=\frac1{\cos^2x}\\\cos^2x=\frac1{1+tg^2x}=\frac1{1+\frac{25}{144}}=\frac1{\frac{169}{144}}=\frac{144}{169}\\\cos x=\sqrt{\frac{144}{169}}=\pm\frac{12}{13}\\x\in[\frac\pi2;\;\pi]\Rightarrow\cos x\ \textless \ 0\\\cos x=-\frac{12}{13}\\\\3)\sin x=\sqrt{1-\cos^2x}=\sqrt{1-\frac{144}{169}}=\sqrt{\frac{25}{169}}=\pm\frac5{13}\\x\in[\frac\pi2;\;\pi]\Rightarrow\sin x\ \textgreater \ 0\\\sin x=\frac5{13}$