Question

Помогоите пожалуйста.

Дано:

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

Answer 1

Формулы тригонометрических функций двойного угла:

$sin2a = frac{2tga}{1+tg^2a}$ 

$cos2a = frac{1-tg^2a}{1+tg^2a}$ 

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

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

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

tg2a = sin2a/cos2a = 1