问题描述: 已知tana=1,3sinB=sin(2a+B),求tan(a+b/2) 1个回答 分类:数学 2014-12-03 问题解答: 我来补答 tana=1 a=kπ+π/4 3sinB=sin(2a+B)=sin(2kπ+π/2+B)=cosB3sinB=cosBsin^2B+cos^2B=1cos^2B=9/10 sin^2B=1/10sinB=√10/10 cosB=3√10/10tan(a+b/2)=(1+tamB/2)/(1-tanB/2)=(cosB/2+sinB/2)/(cosB/2-sinB/2)=(1+sinB)/cosB=(√10+1)/3sinB=-√10/10 cosB=-3√10/10tan(a+b/2)=(1+tamB/2)/(1-tanB/2)=(cosB/2+sinB/2)/(cosB/2-sinB/2)=(1+sinB)/cosB=(-√10+1)/3 展开全文阅读