Билеты по математике для устного экзамена и задачи по теме — страница 2

  • Просмотров 1155
  • Скачиваний 40
  • Размер файла 17
    Кб

sin3xcosx + 1 = sin2x + sinxcos3x tgx - tg2x = sinx 2sin3x - cos2x - sinx = 0 2cos2x = 6(cosx - sinx) 1 - sinx = cosx - sin2x 23sin2(x/2) + 2 = 2sin2x + 3 1 + cos(x2 + 1) = sin2(x2 + 1) 2sinxcos2x + cos4x = 2sinx + cos2x + cos2x tg2x + ctg2x + 3tgx + 3ctgx +4 = 0 1 + cos(x/2) + cosx = 0 1 - sin(x/2) = cosx 2sin2x + cos4x = 0 sin4x + 2cos2x = 1 5sinx - 4ctgx = 0 3cosx + 2tgx = 0 1 + 4cosx = cos2x 2cos2x + 5sinx + 1 = 0 cos2x + 32sinx - 3 = 0 2cos2x + 4cosx =sin2x 2cos2x + sin3x = 2 cos4x + 4sin2x = 1 + 2sin22x 4 - 6cosx = 3 sin2x - sin2(x/2) 5 + 2sin2x - 5cosx = 5sinx cos4x + 8sin2x - 2 = 6cos2x - 8 cos4x 4 - 3cos4x = 10sinxcosx sin4x = (1 +2)(sin2x + cos2x - 1) cos(10x + 12) + 42sin(5x + 6) = 4 sin3x + cos3x = 1 - 1/2sin2x ctg2x - tg2x = 16cos2x 1 +

sinx + cosx + sin2x + cos2x = 0 1/2(cos2x + cos22x) - 1 = 2sin2x - 2sinx - sinx - sin2x tg(/2cosx) = ctg(/2sinx) sin3x - sinx + cos2x = 1 2cos2x + 3sinx = 0 2sin2x + 1/cos2x = 3 2sin2x + 3cosx = 0 1 + sinx+ cosx = 0 sin4x + cos4x = sin2x 4cos4x + 6sin22x + 5cos2x = 0 cos2x + 4sin3x = 1 1 - sin2x = -(sinx + cosx) 4sin22x - 2cos22x = cos8x 8sin4x + 13cos2x = 7 2sinx + 3sin2x = 0 cos(x/2) = 1 + cosx sin2x = 1 + 2cosx + cos2x sin2x = 3sinx 2cos23x - cos3x = 0 3sin2x = 2cos2x 3sin2x - cos2x - 1 = 0 3sin2x - cos2x = 3 Доказать: tg208<sin492 Что больше: sin1 или cos1 tg1 или tg2