Question

sin a = - 1/3; p < a < 3p/2; cos^2 a/2 = ?

Answer 1

$\pi <\alpha <\frac{3\pi }{2} \\\\Sin\alpha=-\frac{1}{3}\\\\Cos^{2}\frac{\alpha }{2}=\frac{1+Cos\alpha}{2}$

α - угол третьей четверти , ⇒  Cosα < 0

$Cos\alpha=-\sqrt{1-Sin^{2}\alpha}=-\sqrt{1-(-\frac{1}{3})^{2}}=-\sqrt{1-\frac{1}{9}}=-\sqrt{\frac{8}{9}}=-\frac{2\sqrt{2}}{3}\\\\Cos^{2}\frac{\alpha }{2}=\frac{1-\frac{2\sqrt{2}}{3} }{2}=\frac{3-2\sqrt{2}}{6}$