【cos(a-π/2)×sin(a-2π)×cos(2π-a)】/sin(5π/2+a)
问题描述:
【cos(a-π/2)×sin(a-2π)×cos(2π-a)】/sin(5π/2+a)
答
解
原式
=[cos(π/2-a)sinacos(-a)]/sin(π/2+a)
=(sinasinacosa)/(cosa)
=sin²a