How do you evaluate $sec(18pi)$ ?

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Answer 1 · 2016-04-24T17:42:28

$sec(18pi)=1$

$sec(18pi)=1/(cos(18pi))$

Since the cosine graph is oscillatory and cyclic with period $2pi$ as shown by the graph below, it implies that $cos(npi)=(-1)^n AAn in NN$ .

Hence $cos(18pi)=(-1)^18=1$

From this it follows that $sec(18pi)=1/1=1$ .

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

How do you evaluate sec(18pi)sec(18pi)?