Вычислить:sin4a, если ctg2a=-2
sin(4α) = 2sin(2α)cos(2α) = [cos(2α)/sin(2α)] * (2sin²(2α)) = 2sin²(2α) * ctg(2α). 1 + ctg²(2α) = (cos²(2α) + sin²(2α)) / sin²(2α) = 1/sin²(2α)); тогда sin(4α) = 2 ctg(2α) / (1 + ctg²(2α)) = 2*(−2) / (1+4) = −4/5.
ответ: −4/5.
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sin(4α) = 2sin(2α)cos(2α) = [cos(2α)/sin(2α)] * (2sin²(2α)) = 2sin²(2α) * ctg(2α).
1 + ctg²(2α) = (cos²(2α) + sin²(2α)) / sin²(2α) = 1/sin²(2α));
тогда
sin(4α) = 2 ctg(2α) / (1 + ctg²(2α)) = 2*(−2) / (1+4) = −4/5.
ответ: −4/5.