Question

How do you determine if `f(x) = 2 cot x` is an even or odd function?

Answer 1

It is odd.

Explanation:

A visual approach is to consider what the graph looks like. An even function is symmetrical over the y-axis. `y = CotX` is not even.
An odd function has point symmetry about the origin (0,0). If you rotate the graph of `y=CotX` about (0,0), it matches the original graph. So it is an odd function.

The algebraic method is to consider that an even function satisfies the condition `f(x) = f(-x)`. Simply substituting x = 30 deg into your calculator will show that `2Cot(30) = 3.464` while `2Cot(-30) = -3.464` which are not the same.

An odd function must satisfy `f(-x) = -f(x)`, which the value for 30 deg clearly satisfies.

To prove that `2Cot(-x) = -2Cot(x)`, the condition for it to be an odd function, we use the trigonometric identities
`Cot(x) =(Cos(x))/(Sin(x)`
`Cos(-x) = Cos(x)`
`Sin(-x) = - Sin(x)`

Then,

`2Cot(-x) = 2((cos(-x))/sin(-x))`
`2cot(-x) = 2((cos(x))/(-sin(x)))`
`2cot(-x) = -2((cos(x))/sin(x))`
`2cot(-x) = -2cot(x)`
Q.E.D.