Question

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

Answer 1

`sec(18pi)=1`

Explanation:

`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]}