神马作文网(1+sin4θ-cos4θ)/(1+sin4θ+cos4θ)-(1+sin4θ+cos4θ)/(1+sin4θ-cos
来源：学生作业帮 编辑：神马作文网作业帮 分类：数学作业 时间：2024/11/12 15:48:12
(1+sin4θ-cos4θ)/(1+sin4θ+cos4θ)-(1+sin4θ+cos4θ)/(1+sin4θ-cos4θ)
(1+sin4θ-cos4θ)/(1+sin4θ+cos4θ)-(1+sin4θ+cos4θ)/(1+sin4θ-cos4θ)
=(1+2sin2θcos2θ-1+2sin²2θ)/(1+2sin2θcos2θ+2cos²2θ-1)-(1+2sin2θcos2θ+2cos²2θ-1)/(1+2sin2θcos2θ-1+2sin²2θ)
=(sin2θcos2θ+sin²2θ)/(sin2θcos2θ+cos²2θ)-(sin2θcos2θ+cos²2θ)/(sin2θcos2θ+sin²2θ)
=sin2θ/cos2θ-cos2θ/sin2θ
=(sin²2θ-cos²2θ)/(sin2θcos2θ)
=-cos4θ/(1/2sin4θ)
=-2cos4θ/sin4θ
=-2cot4θ
=(1+2sin2θcos2θ-1+2sin²2θ)/(1+2sin2θcos2θ+2cos²2θ-1)-(1+2sin2θcos2θ+2cos²2θ-1)/(1+2sin2θcos2θ-1+2sin²2θ)
=(sin2θcos2θ+sin²2θ)/(sin2θcos2θ+cos²2θ)-(sin2θcos2θ+cos²2θ)/(sin2θcos2θ+sin²2θ)
=sin2θ/cos2θ-cos2θ/sin2θ
=(sin²2θ-cos²2θ)/(sin2θcos2θ)
=-cos4θ/(1/2sin4θ)
=-2cos4θ/sin4θ
=-2cot4θ
怎样化简:1-sin4θ-cos4θ/1+sin4θ-cos4θ
1+cos4θ+sin4θ/1-cos4θ+sin4θ可化简为
求证：1+sin4θ−cos4θ2tanθ
求证2cos2θ+sin4θ=cos4θ+1
sin4θ+cos4θ怎样化简?
sin4次方θ+cos4次方θ=1,sinθ-cosθ=
已知tan(Θ+π/4）=3,求1+sin4Θ-cos4Θ/1+sin4Θ+cos4Θ的值.
sin4次方θ-cos4次方θ=5/9 求cos4θ
帮忙化简{根号（1-SIN4*COS4)}
求证1+sin4θ-cos4θ 1+sin4θ+cos4θ——————= ———————— 2tanθ 1-tan
化简1+sin4α+cos4α/1+sin4α -cos4α
化简(1-cos4α+sin4α)/(1+cos4α+sin4α)