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