-1/16
1) Considering that we have the following identity:
[tex]\begin{gathered} \cos (2x)=2\cos (2x)-1 \\ \end{gathered}[/tex]
2) We can plug into that the cos (x).:
[tex]\begin{gathered} \cos (2x)=(\frac{\sqrt[]{15}}{4})^2-1 \\ \cos (2x)=\frac{15}{16}-\frac{16}{15} \\ \cos (2x)=-\frac{1}{16} \end{gathered}[/tex]
Since for the Quadrant II cosine yields a negative value.
