计算：sinπ／12 sin5π／12
1个回答
展开全部
原式=sin(π/4-π/6)sin(π/4+sinπ/6)
=(sinπ/4cosπ/6-cosπ/4sinπ/6)(sinπ/4cosπ/6+cosπ/4sinπ/6)
=(sinπ/4cosπ/6)²-(cosπ/4sinπ/6)²
=(√6/4)²-(√2/4)²
=6/16-2/16
=1/4
=(sinπ/4cosπ/6-cosπ/4sinπ/6)(sinπ/4cosπ/6+cosπ/4sinπ/6)
=(sinπ/4cosπ/6)²-(cosπ/4sinπ/6)²
=(√6/4)²-(√2/4)²
=6/16-2/16
=1/4
本回答由提问者推荐
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
