题目

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

提问时间：2020-10-19

答案

[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