# What does it mean when it says tan is an odd function

1. Jul 12, 2009

What criteria decides whether a trig function is an odd or even?

2. Jul 12, 2009

### Hootenanny

Staff Emeritus
The same criterion as any other odd function: f(-x) = -f(x).

3. Jul 12, 2009

### Staff: Mentor

Graphically speaking, an even function is its own reflection across the y-axis, which makes f(-x) = f(x). An odd function is its own reflection around the origin. This type of reflection is equivalent to a reflection across the x-axis, and then a reflection across the y-axis (or vice versa). This means that if you take, for example, the graph of y = tan x for x > 0, and reflect it across the x-axis, and then the y-axis, it will superimpose exactly on the the half of the graph of y = tan x for x < 0.

4. Jul 12, 2009

I understood about 30% of this.

5. Jul 12, 2009

### Staff: Mentor

Which is the 30% you understand?

6. Jul 12, 2009

What I don't understand is

What is a "reflection"?

What is the difference between reflection across an axis and reflection around the origin?

What it means to reflect the graph?

7. Jul 12, 2009

### snipez90

8. Jul 13, 2009

### Staff: Mentor

Reflection across an axis is pretty straightforward. If a function is its own reflection across the y-axis, that means that for each point (x, y) there is a point on the other side of the y-axis at (-x, y). (I'm assuming that x is positive.) For example, the graph of y = x2 is its own reflection across the y-axis, and is an even function. The relection business is as if a mirror were placed along the y-axis. The part of the graph in the first quadrant has a mirror image in the second quadrant.

Reflection across the origin is less straightforward, since the analogy of mirror images doesn't apply. As I said earlier, an odd function is its own reflection around the origin, which is equivalent to two reflections: one across the x-axis, and the second across the y-axis. For such a function--an odd function--any point (x, y) will have a counterpart at (-x, -y). These points are on a straight line through the origin, and are equidistant from it.

Hope that helps.

9. Jul 13, 2009

### HallsofIvy

But the basic answer to your question is that the definition of "odd function" is that f is an odd function if and only if, for all x for which f is defined, f(-x)= -f(x). Although you didn't ask about it, a f is an "even function" if and only if, for all x for which f is defined, f(-x)= f(x).

The names come from polynomials: If a polynomial has only even powers, say $f(x)= x^6+ 5x^4+ 3x^2+ 2$, then $f(-x)= (-x)^6+ 5(-x)^4+ 3(-x)^2+ 2= x^6+ 5x^4+ 3x^2+ 2= f(x)$ because -x to an even power is x to that power and so f is an even function. Conversely, if f has only off powers: $f(x)= x^5+ 2x^3+ 3x$ then $f(x)= (-x)^5+ 2(-x)^3+ 3(-x)= -x^5- 2x^3- 3x= -(x^5+ 2x^3+ 3x)= -f(x)$ and f is an even function.

But be careful. While all integers are either even or odd, most function are neither. For example, any polynomial that has both even and odd powers is neither an even nor an odd function. ex is also neither even nor odd.

sin(-x)= -sin(x) and cos(-x)= cos(x) so sine is an odd function and cos(x) is an even function. Now tan(x)= sin(x)/cos(x) so tan(-x)= sin(-x)/cos(-x)= what?

Also, we can "separate" any function into its even and odd parts: $f_e(x)= (1/2)(f(x)+ f(-x))$ and $f_o(x)= (1/2)(f(x)+ f(-x))$. $f_e$ is an even function, $f_o$ is an odd function and $f_e(x)+ f_o(x)= f(x)$.

I said that ex is neither even nor odd. Applying those definitions to ex, its even part is $(1/2)(e^x+ e^{-x})= cosh(x)$ and its odd part is $(1/2)(e^x- e^{-x})= sinh(x)$.