How does one find sin2x, cos2x, and tan2x if sin x = -2/#sqrt13# and x is in quadrant IV?

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Answer 1 · 2018-04-27T05:01:02

Please see below.

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#sinx=-2/sqrt13#

In quadrant #IV# #sinx < 0# as indicated. But #cosx > 0#.

#cosx=sqrt(1-sin^2x)=sqrt(1-4/13)=sqrt(9/13)=3/sqrt13#

#sin2x=2sinxcosx=2(-2/sqrt13)(3/sqrt13)=-12/13#

#cos2x=2cos^2x-1=2(9/13)-1=18/13-1=5/13#

#tan2x=(sin2x)/(cos2x)=(-12/13)/(5/13)=-12/5#