Tgα = √2 +1
sinα = tgα/[+/-√[1+tg²α)]
cosα = 1/[+/-√(1+tg²α)] ⇒
sin2α = 2·sinα · cosα = 2·tgα/(1+tg²α) ⇒
sin2α = 2 · (√2+1)/[1 +(√2+1)²] = 2 · (√2+1)/(1 +2+2√2 +1) =
= 2 · (√2+1)/ (4+2√2) = (√2+1)/(2+√2)+
= [(√2+1)·(2-√2)]/[(2+√2)·(2 - √2)] =
=(2√2+2 - 2 - √2)/(4 - 2) = 2·(√2 - 1)/2 = √2 -1
Ответ: sin2α = √2 - 1