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let sinA =3/5 in QI and sinB =-5/13 in QIV find sin2A, cos2B, and tan2A
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use sin2A=2sinAcosA cos2B=cos^2B-sin^2B tan2A=2tanA/(1-tan^2A)
For tan2A, you can also use, \[\tan(2A)=\sin (2A)/\cos(2A)\]
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