若sin(180° α)=-1010,0°<α<90°.求sin(-α) sin(-90°-α)cos(540°-α) cos(-270°-α)的值.
试题难度:难度:中档 试题类型:解答题 试题内容:若sin(180° α)=-1010,0°<α<90°.求sin(-α) sin(-90°-α)cos(540°-α) cos(-270°-α)的值.
试题答案:由sin(180° α)=-1010,α∈(0°,90°),
得sin α=1010,cos α=31010,
∴原式=-sinα-sin(90° α)cos(360° 180°-α) cos(270° α)
=-sinα-cosα-cosα sinα
=-1010-31010-31010 1010
=2.
答:试题答案:18×9 3=162 5,165,答:A=165.故答案为:165.详情>>
答:详情>>