Question

三角形ABC中,cosA=5/13,tanB/2+cotB/2=10/3,求cos(A-B)的值

Answer 1

tanB/2+cotB/2=10/3

(sinB/2)/(cosB/2)+(cosB/2)/(sinB/2)=10/3

[(sinB/2)^2+(cosB/2)^2]/[(cosB/2)(sinB/2)]=10/3

1/[(sinB)/2]=10/3

sinB=3/5

因(sinB)^2+(cosB)^2=1所以cosB=4/5,

同理,sinA=12/13

cos(A-B)

=cosA*cosB+sinA*sinB

=(5/13)*(4/5)+(12/13)*(3/5)

=56/65