【cos(a-π/2)×sin(a-2π)×cos(2π-a)】/sin(5π/2+a)

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/06 08:09:46

【cos(a-π/2)×sin(a-2π)×cos(2π-a)】/sin(5π/2+a)
【cos(a-π/2)×sin(a-2π)×cos(2π-a)】/sin(5π/2+a)

【cos(a-π/2)×sin(a-2π)×cos(2π-a)】/sin(5π/2+a)

原式
=[cos(π/2-a)sinacos(-a)]/sin(π/2+a)
=(sinasinacosa)/(cosa)
=sin²a