Question

sin a= -5/13, п найдите cos, tg, ctg

Answer 1

$\displaystyle\bf\\\pi < \alpha <\frac{3\pi }{2} \ \ \Rightarrow \ \ Cos\alpha <0 \ , \ tg\alpha >0 \ , \ Ctg\alpha >0\\\\\\Cos\alpha =-\sqrt{1-Sin^{2} \alpha } =-\sqrt{1-\Big(-\frac{5}{13} \Big)^{2} } =-\sqrt{1-\frac{25}{169} } =\\\\\\=-\sqrt{\frac{144}{169} } =-\frac{12}{13} \\\\\\tg\alpha =\frac{Sin\alpha }{Cos\alpha } =\Big(-\frac{5}{13} \Big):\Big(-\frac{12}{13} \Big)=\frac{5}{13} \cdot\frac{13}{12} =\frac{5}{12}$

$\displaystyle\bf\\Ctg\alpha =\frac{1}{tg\alpha } =1:\frac{5}{12} =1\cdot\frac{12}{5} =\frac{12}{5} =2,4$