Question

Помогите с 5 заданием,пожалуйста

Answer 1

Решение
sinα = 1/3
cosα = √(1 - sin²α) = √(1 - 1/9) = √(8/9) = 2√2 / 3
tgα = sinα/cosα = 1/3 : 2√2/3 = (1*3)/(3*2√2) = √2/4
tg2α = 2tgα / (1 - tg²α) = [2*(√2/4)] / [1 - (√2/4)²] = (√2/2) : 7/8 =
= (√2 * 8) / (2*7) = 4√2 / 7
ctg2α = 1 / tg2α
ctg²2α = 1 /  tg²2α
ctg²2α = 1 /[4√2)/7]² = 49/32