Question

Help me please i want to see if i did it correctly

Answer 1

-1/16

1) Considering that we have the following identity:

`\begin{gathered} \cos (2x)=2\cos (2x)-1 \\ \end{gathered}`

2) We can plug into that the cos (x).:

`\begin{gathered} \cos (2x)=(\frac{\sqrt[]{15}}{4})^2-1 \\ \cos (2x)=(15)/(16)-(16)/(15) \\ \cos (2x)=-(1)/(16) \end{gathered}`

Since for the Quadrant II cosine yields a negative value.