Question

срочно нужен ответ. помогите пожалуйста готов дать 50 баллов​

Answer 1

β - угол второй четверти значит Cosβ < 0 .

$\displaystyle\bf\\Sin\beta =\frac{3}{5} \\\\\\Cos\beta =-\sqrt{1-Sin^{2} \beta } =-\sqrt{1-\Big(\frac{3}{5}\Big)^{2} } =-\sqrt{1-\frac{9}{25} } =\\\\\\=-\sqrt{\frac{16}{25} } =-\frac{4}{5} \\\\\\Sin2\beta =2Sin\beta Cos\beta =2\cdot\frac{3}{5} \cdot\Big(-\frac{4}{5} \Big)=-\frac{24}{25} \\\\\\\boxed{Sin2\beta =-\frac{24}{25}}\\\\\\Cos2\beta =1-2Sin^{2} \beta =1-2\cdot\Big(\frac{3}{5} \Big)^{2} =1-2\cdot\frac{9}{25} =\\\\\\=1-\frac{18}{25}=\frac{7}{25}$

$\displaystyle\bf\\\boxed{Cos2\beta =\frac{7}{25}}\\\\\\tg2\beta =\frac{Sin2\beta }{Cos2\beta } =-\frac{24}{25} :\frac{7}{25} =-\frac{24}{25} \cdot\frac{25}{7} =-\frac{24}{7} =-3\frac{3}{7} \\\\\\\boxed{tg2\beta =-3\frac{3}{7} }\\\\\\Ctg2\beta =\frac{1}{tg2\beta } =1:\Big(-\frac{24}{7} \Big)=-1\cdot\frac{7}{24} =-\frac{7}{24} \\\\\\\boxed{Ctg2\beta =-\frac{7}{24} }$