题目

已知tana=1,3sinB=sin(2a+B),求tan(a+b/2)

提问时间：2020-10-07

答案

tana=1 a=kπ+π/4
3sinB=sin(2a+B)=sin(2kπ+π/2+B)=cosB
3sinB=cosB
sin^2B+cos^2B=1
cos^2B=9/10 sin^2B=1/10
sinB=√10/10 cosB=3√10/10
tan(a+b/2)=(1+tamB/2)/(1-tanB/2)=(cosB/2+sinB/2)/(cosB/2-sinB/2)
=(1+sinB)/cosB
=(√10+1)/3
sinB=-√10/10 cosB=-3√10/10
tan(a+b/2)=(1+tamB/2)/(1-tanB/2)=(cosB/2+sinB/2)/(cosB/2-sinB/2)
=(1+sinB)/cosB
=(-√10+1)/3