Question

let sinA =3/5 in QI and sinB =-5/13 in QIV find sin2A, cos2B, and tan2A

Answer 1

use
sin2A=2sinAcosA
cos2B=cos^2B-sin^2B
tan2A=2tanA/(1-tan^2A)

Answer 2

For tan2A, you can also use,
\[\tan(2A)=\sin (2A)/\cos(2A)\]

Answer 3

yes :)