F(x)=1/(2sin3x) 2(sin2x+cos2x)/cosx-sinx-cos3x+sin3x - вопрос №673663712322233 от sanya1897p08pyj 05.05.2020 14:23

Question

Алгебра: F(x)=1/(2sin3x) 2(sin2x+cos2x)/cosx-sinx-cos3x+sin3x

sanya1897p08pyj · Answer

sin2x + cos2x = 1tgx  =  sinxcosxctgx  =  cosxsinxtgx ctgx = 1tg2x + 1  =  1cos2xctg2x + 1  =  1sin2xФормулы двойного аргументаsin2x = 2sinx cosxsin2x  =  2tgx  = 2ctgx  = 21 + tg2x1 + ctg2xtgx + ctgxcos2x = cos2x - sin2x = 2cos2x - 1 = 1 - 2sin2xcos2x  =  1 - tg2x  = ctg2x - 1  = ctgx - tgx1 + tg2xctg2x + 1ctgx + tgxtg2x  =  2tgx  = 2ctgx  = 21 - tg2xctg2x - 1ctgx - tgxctg2x  =  ctg2x - 1  = ctgx - tgx2ctgx2...