Given `sin(u)=-3/5, (3pi)/2<u<2pi`
Angle u is in quadrant 4. A right triangle is drawn in quadrant 4. Since`sin(u)=-3/5`
the side opposite from angle u is 3 and the hypotenuse is 5. Using the pythagorean theorem, the side adjacent to angle u is 4.
`sin(2u)=2sin(u)cos(u)=2(-3/5)(4/5)=-24/25`
`cos(2u)=1-2sin^2(u)=1-2(-3/5)^2=1-2(9/25)=25/25-18/25=7/25`
`tan(2u)=2tan(u)/(1-tan^2(u))=2(-3/4)/(1-(-3/4)^2)=(-3/2)/(1-9/16)=(-3/2)/(16/16-9/16)=(-3/2)/(7/16)=-24/7`
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