来自杜海宁的问题
高一三角函数化简问题,要过程,急!tan(2π-α)sin(-2π-α)cos(6π-α)/cos(α-π)sin(5π+α)
高一三角函数化简问题,要过程,急!
tan(2π-α)sin(-2π-α)cos(6π-α)/cos(α-π)sin(5π+α)
1回答
2020-08-02 14:19
高一三角函数化简问题,要过程,急!tan(2π-α)sin(-2π-α)cos(6π-α)/cos(α-π)sin(5π+α)
高一三角函数化简问题,要过程,急!
tan(2π-α)sin(-2π-α)cos(6π-α)/cos(α-π)sin(5π+α)
tan(2π-α)sin(-2π-α)cos(6π-α)/cos(α-π)sin(5π+α)
=tan(-α)sin(-α)cos(-α)/cos(π-α)sin(π+α)
=(sin(-α))^2/(-cosα)(-sinα)
=(sinα)^2/(cosα)(sinα)
=(sinα)/(cosα)
=tanα
完毕.其中第二步因为tan(-α)=sin(-α)/cos(-α).