Question

sin a, tg a, ctg a, якщо cos a = -1/4 і 90° < a < 180°​

Answer 1

$\displaystyle\bf\\90^\circ < \alpha < 180^\circ \ \ \Rightarrow \ \ Sin\alpha > 0 \ ; \ tg\alpha < 0 \ , \ Ctg\alpha < 0\\\\\\Cos\alpha =-\frac{1}{4} \\\\\\Sin\alpha =\sqrt{1-Cos^{2} \alpha } =\sqrt{1-\Big(-\frac{1}{4} \Big)^{2} } =\sqrt{1-\frac{1}{16} } =\sqrt{\frac{15}{16} } =\frac{\sqrt{15} }{4}\\\\\\tg\alpha =\frac{Sin\alpha }{Cos\alpha } =\frac{\sqrt{15} }{4} :\Big(-\frac{1}{4} \Big)=-\frac{\sqrt{15} }{4} \cdot 4=-\sqrt{15}$

$\displaystyle\bf\\Ctg\alpha =\frac{1}{tg\alpha } =\frac{1}{-\sqrt{15} } =-\frac{\sqrt{15} }{15}$