作业帮求证:[sin^2(π+α)*cos(π/2-α)+tan(2π-α)*cos(-α)]/-sin^2(-α)+tan(
来源：学生作业帮 编辑：作业帮 分类：数学作业 时间：2024/05/07 18:53:45
求证:[sin^2(π+α)*cos(π/2-α)+tan(2π-α)*cos(-α)]/-sin^2(-α)+tan(-π+α)cot(α-π)=-sinα
:[sin^2(π+α)*cos(π/2-α)+tan(2π-α)*cos(-α)]/[-sin^2(-α)+tan(-π+α)cot(α-π)]=-sinα
:[sin^2(π+α)*cos(π/2-α)+tan(2π-α)*cos(-α)]/[-sin^2(-α)+tan(-π+α)cot(α-π)]=-sinα
[sin^2(π+α)*cos(π/2-α)+tan(2π-α)*cos(-α)]/[-sin^2(-α)+tan(-π+α)cot(α-π)]
=[sin^2a*sina+(-tana)cosa] / [-sin^2a+tana*cota]
=(sin^3a-sina) / (1-sin^2a)
=sina(sin^2a-1)/(1-sin^2a)
=-sina
=[sin^2a*sina+(-tana)cosa] / [-sin^2a+tana*cota]
=(sin^3a-sina) / (1-sin^2a)
=sina(sin^2a-1)/(1-sin^2a)
=-sina
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