Question

известно, что cos a = 1/9,π<a<2π
найдите sin a/2, cos a/2 и tg a/2​

Answer 1

$\displaystyle\bf\\\pi < \alpha < 2\pi \ \ \ \Rightarrow \ \ \ \frac{\pi }{2} < \frac{\alpha }{2} < \pi \ \ \ \Rightarrow \ \ Sin\frac{\alpha }{2} > 0 \ , \ Cos\frac{\alpha }{2} < 0 \ ,tg\frac{\alpha }{2} < 0\\\\\\Sin^{2} \frac{\alpha }{2} =\frac{1-Cos\alpha }{2} =\frac{1-\dfrac{1}{9} }{2} =\frac{8}{9} :2=\frac{4}{9} \\\\\\Sin\frac{\alpha }{2} =\sqrt{\frac{4}{9} } =\frac{2}{3}$

$\displaystyle\bf\\Cos^{2} \frac{\alpha }{2} =\frac{1+Cos\alpha }{2} =\frac{1+\dfrac{1}{9} }{2} =\frac{10}{9} :2=\frac{5}{9} \\\\\\Cos\frac{\alpha }{2} =-\sqrt{\frac{5}{9} } =-\frac{\sqrt{5} }{3}\\\\\\tg\frac{\alpha }{2} =\frac{Sin\frac{\alpha }{2} }{Cos\frac{\alpha }{2} } =\frac{2}{3} :\Big(-\frac{\sqrt{5} }{3} \Big)=-\frac{2}{3}\cdot\frac{3}{\sqrt{5} } =-\frac{2}{\sqrt{5} } =-\frac{2\sqrt{5} }{5}$

$\displaystyle\bf\\Otvet:Sin\frac{\alpha }{2} =\frac{2}{3} \ \ ; \ \ Cos\frac{\alpha }{2} =-\frac{\sqrt{5} }{3} \ \ ; \ \ tg\frac{\alpha }{2} =-\frac{2\sqrt{5} }{5}$