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Ответил dnepr1
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а) cos β + cos((2π/3)+β) + cos((2π/3)-β) =
cos β + cos(2π/3)*cos β + sin(2π/3)*sin β + cos(2π/3)*cos β - sin(2π/3)*sin β =
= cos β + (-1/2)*cos β + (√2/3)*sin β + (-1/2)*cos β - (√2/3)*sin β = = 0.
б) ((2tgα) / (1+tg²α))*2cos(2α) = tg(2α)*2cos(2α) =
= sin(2α)*2cos(2α) / cos(2α) = 2 sin(2α).
cos β + cos(2π/3)*cos β + sin(2π/3)*sin β + cos(2π/3)*cos β - sin(2π/3)*sin β =
= cos β + (-1/2)*cos β + (√2/3)*sin β + (-1/2)*cos β - (√2/3)*sin β = = 0.
б) ((2tgα) / (1+tg²α))*2cos(2α) = tg(2α)*2cos(2α) =
= sin(2α)*2cos(2α) / cos(2α) = 2 sin(2α).
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