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

1 Answer
Dec 3, 2016

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.