Вопрос от пользователя: Селена
F (x) = 1 / (2sin3x)
2 (sin2x+cos2x) / cosx-sinx-cos3x+sin3x
Ответ: Лексей
sin2x + cos2x = 1
tgx = sinx cosx ctgx = cosx sinx
tgx ctgx = 1
tg2x + 1 = 1 cos2x ctg2x + 1 = 1 sin2x Формулы двойного аргумента
sin2x = 2sinx cosx
sin2x = 2tgx = 2ctgx = 2 1 + tg2x 1 + ctg2x tgx + ctgx
cos2x = cos2x — sin2x = 2cos2x — 1 = 1 — 2sin2x
cos2x = 1 — tg2x = ctg2x — 1 = ctgx — tgx 1 + tg2x ctg2x + 1 ctgx + tgx tg2x = 2tgx = 2ctgx = 2 1 — tg2x ctg2x — 1 ctgx — tgx ctg2x = ctg2x — 1 = ctgx — tgx 2ctgx 2
