问题描述： cos(a-β/2)=-1/9,sin(a/2-β）=2/3 且 π/2 1个回答 分类：数学 2014-09-22 问题解答： 我来补答 π/2＜a＜π,0＜b＜π/2 --->π/4＜a-b/2＜π,∵cos(a-b/2)=-1/9＜0--->π/2＜a-b/2＜π --->sin(a-b/2)=4√5/9 同时：-π/4＜a/2-b＜π/2,∵sin(a/2-b)＞0--->0＜a/2-b＜π/2 --->cos(a/2-b)=√5/3 cos(a/2+b/2) = cos[(a-b/2)-(a/2-b)] = cos(a-b/2)cos(a/2-b)+sin(a-b/2)sin(a/2-b) = (-1/9)(√5/3)+(4√5/9)(2/3) = 7√5/27 展开全文阅读
