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化简(1+sin4x+cos4x)/(1+sin4x-cos4x)
人气:256 ℃ 时间:2020-04-03 02:33:08
解答
原式=(sin2x^2+cos2x^2+2sin2xcos2x+cos2x^2-sin2x^2)/(sin2x^2+cos2x^2+2sin2xcos2x-cos2x^2+sin2x^2)
=(2cos2x^2+2sin2xcos2x)/(2sin2x^2+2sin2xcos2x)
=[2(sin2x+cos2x)cos2x]/[2(sin2x+cos2x)sin2x]
=cos2x/sin2x=ctg2x
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