Question

已知tana=1,3sinB=sin(2a+B),求tan(a+b/2)

Answer 1

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