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

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We need to compute $\cos (\dfrac{-3 \pi}{2})$. The trigonometric function $\cos$ has a period of $2\pi$. Also, because cosine is even, $\cos (-x)=\cos (x)$ Thus, we can simplify as follows: $\cos (\dfrac{-3 \pi}{2}) =\cos (\dfrac{3 \pi}{2}) \\ =0$