ПОЖАЛЙСТА!!!с дано:cost=34; 0 Ответы: 06-06-2017 07:44 cost = 3/4 cos(t/2) = ((1 + cost)/2)^(1/2) = ((1 + 3/4)/2)^(1/2) = (7/8)^(1/2) = (7^(1/2))/(2*(2^(1/2))) sin(t/2) = (1 - cos^2(t/2))^(1/2) = (1 - 7/8)^(1/2) = 1/(8^(1/2)) = 1/(2*(2^(1/2))) tg(t/2) = sin(t/2)/cos(t/2) = (1/(2*(2^(1/2))))/((7^(1/2))/(2*(2^(1/2)))) = 1/(7^(1/2)) ctg(t/2) = 1/tg(t/2) = 7^(1/2)
Ответы:
06-06-2017 07:44
cost = 3/4 cos(t/2) = ((1 + cost)/2)^(1/2) = ((1 + 3/4)/2)^(1/2) = (7/8)^(1/2) = (7^(1/2))/(2*(2^(1/2))) sin(t/2) = (1 - cos^2(t/2))^(1/2) = (1 - 7/8)^(1/2) = 1/(8^(1/2)) = 1/(2*(2^(1/2))) tg(t/2) = sin(t/2)/cos(t/2) = (1/(2*(2^(1/2))))/((7^(1/2))/(2*(2^(1/2)))) = 1/(7^(1/2)) ctg(t/2) = 1/tg(t/2) = 7^(1/2)
⭐⭐⭐⭐⭐ Лучший ответ на вопрос «ПОЖАЛЙСТА!!!с дано:cost=34; 0
Открой этот вопрос на телефоне - включи камеру и наведи на QR-код!