## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$\dfrac{- 3}{5}$
We have: $\sin \theta=\dfrac{2}{\sqrt 5}$ and $\cos \theta=-\dfrac{1}{\sqrt 5}$ We plug these values into the following identity: $\cos (2 \ \theta)=\cos^2 \theta -\sin^2 \theta = (\dfrac{-1}{\sqrt 5})^2 -(\dfrac{2}{\sqrt 5})^2=\dfrac{- 3}{5}$