# How do you evaluate sec(18pi)?

Apr 24, 2016

$\sec \left(18 \pi\right) = 1$

#### Explanation:

$\sec \left(18 \pi\right) = \frac{1}{\cos \left(18 \pi\right)}$

Since the cosine graph is oscillatory and cyclic with period $2 \pi$ as shown by the graph below, it implies that $\cos \left(n \pi\right) = {\left(- 1\right)}^{n} \forall n \in \mathbb{N}$.

Hence $\cos \left(18 \pi\right) = {\left(- 1\right)}^{18} = 1$

From this it follows that $\sec \left(18 \pi\right) = \frac{1}{1} = 1$.

graph{cosx [-7.9, 7.9, -3.95, 3.95]}