Question

Даю 50 баллов
cosα=-4/5; π<α<3π/2
Найти
sin α/2=?
cos α/2=?
tg α/2=?

Answer 1

$cos\alpha =-\dfrac{4}{5}\\\\\\sin^2\dfrac{\alpha }{2}=\dfrac{1-cos\alpha }{2}=\dfrac{1+\dfrac{4}{5}}{2}=\dfrac{9}{10}\ \ \to \ \ sin\dfrac{\alpha }{2}=\pm \sqrt{\dfrac{9}{10}}=\pm \dfrac{3}{\sqrt{10}}\\\\\\\pi <\alpha <\dfrac{3\pi}{2}\ \ \ \to \ \ \ \dfrac{\pi}{2}<\dfrac{\alpha }{2}<\dfrac{3\pi}{4}\ \ \Rightarrow \ \ sin\dfrac{\alpha }{2}>0\ \ ,\ \ \ sin\dfrac{\alpha }{2}=+\dfrac{3}{\sqrt{10}}$

$\dfrac{\pi}{2}<\dfrac{\alpha }{2}<\dfrac{3\pi}{4}\ \ \to \ \ cos\dfrac{\alpha }{2}<0\ \ \Rightarrow \\\\\\cos\dfrac{\alpha }{2}=-\sqrt{1-sin^2\dfrac{\alpha }{2}}=-\sqrt{1-\dfrac{9}{10}}=-\dfrac{1}{\sqrt{10}}\\\\\\tg\dfrac{\alpha }{2}=\dfrac{sin\frac{\alpha }{2}}{cos\frac{\alpha }{2}}=\dfrac{\frac{3}{\sqrt{10}}}{-\frac{1}{\sqrt{10}}} =-3$