Question

cos a = 3/5
Найдите sin a/2, cos a/2, tg a/2

Answer 1

$cos\alpha =\frac{3}{5}\\\\sin^2\frac{\alpha }{2}=\frac{1-cos\alpha }{2}=\frac{1-\frac{3}{5}}{2}=\frac{2}{5\cdot 2}=\frac{1}{5}\; \; ,\; \; sin\frac{\alpha }{2}=\pm \frac{1}{\sqrt5}=\pm \frac{\sqrt5}{5}\\\\cos^2\frac{\alpha }{2}=\frac{1+cos\alpha }{2}=\frac{1+\frac{3}{5}}{2}=\frac{8}{5\cdot 2}=\frac{4}{5}\; \; ,\; \; cos\frac{\alpha }{2}=\pm \frac{2}{\sqrt5}=\pm \frac{2\sqrt5}{5}\\\\tg\frac{\alpha }{2}=\pm \sqrt{\frac{1-cos\alpha }{1+cos\alpha }}=\pm \sqrt{\frac{1-\frac{3}{5}}{1+\frac{3}{5}}}=\pm \sqrt{\frac{2}{8}}=\pm \frac{1}{2}$