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化简cosA*cos2A*cos4A……cos2^nA
人气:208 ℃ 时间:2020-06-06 13:55:03
解答
cosA*cos2A*cos4A……cos2^nA =sinA*cosA*cos2A*cos4A……cos2^nA/sinA =sin2A*cos2A*cos4A……cos2^nA/2sinA =sin4A……cos2^nA/4sinA =sin2^(n+1)A/[2^(n+1)sinA]
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