Помогоите пожалуйста. Дано: ctgα =√2+1 и 0°<α<45° .Найдите sin2α ,cos2α ,tg2α .

Формулы тригонометрических функций двойного угла:

$sin2a = \frac{2tga}{1+tg^2a}$

$cos2a = \frac{1-tg^2a}{1+tg^2a}$

$tga = 1/ctga = \frac{1}{\sqrt{2}+1}$

$sin2a = \frac{2}{\sqrt{2}+1}:(1+\frac{1}{(\sqrt{2}+1)^2}) = \frac{2}{\sqrt{2}+1}: \frac{2+2\sqrt{2}+1+1}{(\sqrt{2}+1)^2} = \frac{2(\sqrt{2}+1)}{2\sqrt{2}+4} = \frac{\sqrt{2}+1x}{\sqrt{2}+2}$

$cos2a = (1-\frac{1}{(\sqrt{2}+1)^2}):(1+\frac{1}{(\sqrt{2}+1)^2}) = \frac{2+2\sqrt{2}+1-1}{(\sqrt{2}+1)^2}:\frac{2+2\sqrt{2}+1+1}{(\sqrt{2}+1)^2} = \frac{2\sqrt{2}+2}{2\sqrt{2}+4} = \frac{\sqrt{2}+2}{\sqrt{2}+2}$

tg2a = sin2a/cos2a = 1