Ответы на вопрос:
Sin 2α = 2sinαcosα cos²α = 1 - sin²α = 1 - 1/4 = 3/4 cosα = √3/2 sin2α = 2 * 1/2 * √3/2 = √3/2 cos 2α = cos²α - sin²α = 3/4 - 1/4 = 2/4 = 1/2 tg 2α = 2tg α / ( 1 - tg²α) = (2 * 1/√3) / (1 - 1/3) = 2/√3 * 3/2 = 3/√3
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