求证:tan5x+tan3xcos2x•cos4x=4(tan5x−tan3x)
来源:学生作业帮 编辑:百度作业网作业帮 分类:数学作业 时间:2024/07/31 15:42:11
求证:
=4(tan5x−tan3x)
tan5x+tan3x |
cos2x•cos4x |
证明:要证
tan5x+tan3x
cos2x•cos4x=4(tan5x−tan3x)成立
即证
sin5x
cos5x+
sin3x
cos3x
cos2xcos4x=4(
sin5x
cos5x−
sin3x
cos3x)成立,即
sin5xcos3x+cos5xsin3x
cos2xcos4x=4(sin5xcos3x−cos5xsin3x)
即证
sin8x
cos2xcos4x=4sin2x成立
又因为sin8x=2sin4xcos4x=4sin2xcos2xcos4x
所以左边=
sin8x
cos2xcos4x=
4sin2xcos2xcos4x
cos2xcos4x=4sin2x=右边
得证.
tan5x+tan3x
cos2x•cos4x=4(tan5x−tan3x)成立
即证
sin5x
cos5x+
sin3x
cos3x
cos2xcos4x=4(
sin5x
cos5x−
sin3x
cos3x)成立,即
sin5xcos3x+cos5xsin3x
cos2xcos4x=4(sin5xcos3x−cos5xsin3x)
即证
sin8x
cos2xcos4x=4sin2x成立
又因为sin8x=2sin4xcos4x=4sin2xcos2xcos4x
所以左边=
sin8x
cos2xcos4x=
4sin2xcos2xcos4x
cos2xcos4x=4sin2x=右边
得证.
求两道极限题的解法lim(x^6-1)/(x^10-1)x→1lim(tan3x)/(tan5x)x→0答案都是3/5,
求X趋向0时,sin3X/tan5x的极限
计算极限 lim x→0 tan5x/x
lim x→0(tan2x/tan5x)的极限怎么求?
洛必达法则求limx趋于丌,sin3x/tan5x的极限
lim sin3x/tan5x在x->pai时的极限
求极限limx趋于0(ln(1+2x))/tan5x
lim(x→∏) tan5x/sin3x = lim(x→∏) 5x•tan5x/5x / 3x ̶
lim(x趋于π(派)) sin3x/tan5x.lim(x趋于π/4(4分之派)) sinx-cosx/1-cos^x
用洛必达法则求当x→π时(sin3x)/(tan5x)的极限
当x趋向于0时,求(sin3x)\(tan5x)的极限
用洛必达法则求极限lim(X-π)sin3X除以tan5x请求写出详细求解过程谢谢