#### Answer

$sin~2y = -\frac{24}{25}$
$cos~2y = -\frac{7}{25}$

#### Work Step by Step

Since $sec~y$ is negative and $sin~y$ is positive, the angle $y$ lies in quadrant II.
$sec~y = \frac{hyp}{adj} = -\frac{5}{3} = \frac{5}{-3}$
If the hypotenuse is 5 and the adjacent side is -3, then the adjacent side is $\sqrt{(5)^2-(-3)^2} = 4$
$sin~y = \frac{opp}{hyp} = \frac{4}{5}$
$cos~y = \frac{adj}{hyp} = \frac{-3}{5}$
We can find $sin~2y$:
$sin~2y = 2~sin~y~cos~y$
$sin~2y = (2)(\frac{4}{5})(\frac{-3}{5})$
$sin~2y = -\frac{24}{25}$
We can find $cos~2y$:
$cos~2y = cos^2~y-sin^2~y$
$cos~2y = (\frac{-3}{5})^2-(\frac{4}{5})^2$
$cos~2y = \frac{9}{25}-\frac{16}{25}$
$cos~2y = -\frac{7}{25}$