Question

Найти tg 2 x. Если cos x= - 3/5 П < x< 3/2 П *

Answer 1

Ответ:

$cosx=-\dfrac{3}{5}\\\\\sin^2x=1-cos^2x=1-\dfrac{9}{25}=\dfrac{16}{25}\ \ \ \to \ \ \ sinx=\pm \dfrac{4}{5}\\\\\pi <x<\dfrac{3\pi }{2}\ \ \to \ \ sinx<0\ \ \to \ \ sinx=-\dfrac{4}{5}\\\\tgx=\dfrac{sinx}{cosx}=\dfrac{-4/5}{-3/5}=\dfrac{4}{3}\\\\\\tg2x=\dfrac{2\, tgx}{1-tg^2x}=\dfrac{\frac{8}{3}}{1-\frac{16}{9}}=\dfrac{8}{3\cdot (-\frac{7}{9})}=-\dfrac{8\cdot 3}{7}=-\dfrac{24}{7}$