[tex]\sin2u=2\sin u\cos u[/tex]
We know that [tex]\cos u=-\dfrac23[/tex], so we can find out what [tex]\sin u[/tex] is, but there are two possibilities. By the Pythagorean theorem,
[tex]\sin^2u+\cos^2u=1\implies \sin u=\pm\sqrt{1-\left(-\dfrac23\right)^2}=\pm\dfrac{\sqrt5}3[/tex]
and so
[tex]\sin2u=2\left(\pm\dfrac{\sqrt5}3\right)\left(-\dfrac23\right)=\pm\dfrac{4\sqrt5}9[/tex]
Next,
[tex]\cos2u=\cos^2u-\sin^2u[/tex]
and since [tex]x^2\ge0[/tex] for all [tex]x[/tex], we end up with exactly one value for [tex]\cos2u[/tex]:
[tex]\cos2u=\left(-\dfrac23\right)^2-\left(\pm\dfrac{\sqrt5}3\right)^2=-\dfrac19[/tex]
