化简[2sin(2π+2α)/(1+cos2α)]*[sin²(3π/2-α)/cos(π-2α)]
问题描述：
化简[2sin(2π+2α)/(1+cos2α)]*[sin²(3π/2-α)/cos(π-2α)]
高一数学题
实在不好意思...
我看串行了...
答
[2sin(2π+2α)/(1+cos2α)]*[sin²(3π/2-α)/cos(π-2α)]
=-[sin(2α)/(1+cos2α)]*[2cos²(α)/cos(2α)]
=-[sin(2α)/cos(2α)]*[2cos²(α)/(1+cos2α)]
=-tan2α