Question

已知cosα=3/5,cos(α-β)=7√2/10

已知cosα=3/5,cos(α-β)=12/13,且0<β<α<π/2.
①求tan2α的值
②求cosβ

Answer 1

cosα=3/5,cos(α-β)=12/13
0<β<α<π/2
α-β<0
sinα=4/5,sin(α-β)=-5/13
tanα=sinα/cosα
=(4/5)/(3/5)
=4/3
tan2α
=2tanα/(1-tan^2α)
=(2*4/3)/(1-16/9)
=(8/3)/(7/9)
=8/3*9/7
=24/7
cosβ
=cos[α-(α-β)]
=cosαcos(α-β)+sinαsin(α-β)
=3/5*12/13-4/5*5/13
=36/65-20/65
=16/65