Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Assuming I have the graph of a function $f(x)$ is there function $f_1(f(x))$ that will give me a rotated version of the graph of that function?

For example if I plot $\sin(x)$ I will get a sine wave which straddles the $x$-axis, can I apply a function to $\sin(x)$ to yield a wave that straddles the line that would result from $y = 2x$?

share|cite|improve this question
The "graph of a function" here is actually the graph of points satisfying an equation relating $x$ and $y$, for example $y = \sin x.$ A perfectly correct answer that was posted below (and accepted) tells us how to rotate the graph of any equation relating $x$ and $y,$ even if the equation cannot be written in the form $y = f(x).$ For example, you can rotate the graph of $y^2 = 4 - 2x^2,$ which is an ellipse, by this same method. – David K Dec 14 '14 at 21:08
up vote 10 down vote accepted

Once you rotate, it need not remain a function (i.e. one $x$ value can have multiple $y$ values corresponding to it).

But you can use the following transformation

$$x' = x\cos \theta - y \sin \theta$$ $$y' = x \sin \theta + y \cos \theta$$

to rotate by an angle of $\theta$. Point $(x,y)$ gets rotated to point $(x',y')$. Note: this is a rotation about the origin.

In your case of $y = 2x$, you need to rotate by $\arctan(2)$.

See this for more info: Rotation Matrix.

share|cite|improve this answer
In the specific case of sin(x) if you rotate by something in the range (-pi/4,pi/4) you still have a function because the slope of sin(x) is always less than 1 in absolute value. – Ross Millikan Jan 12 '11 at 17:34
I think this is what I want. I want to animate something so that it undulates as if it were following a sin wave but not always in parallel to the x axis. – Omar Kooheji Jan 12 '11 at 17:43
@Omar: It looks like the parameterization that Rasmus refers to will suit you nicely. It gives you x and y each as a function of t and avoids the "double x" problem. – Ross Millikan Jan 12 '11 at 19:25

In general, the answer is no since the rotated version of the graph might not be the graph of a function. For instance it could happen that your rotated version of the graph contains two different points with the same $x$-value -- this cannot happen for the graph of a function.

A way out could be to parametrise your graph. So instead of a map $x\mapsto y(x)$ you look at the map $t\mapsto (t,y(t))$. After rotating the trajectory of this thing (not the graph!) it will still be the trajectory of a map $$t\mapsto (x(t),y(t)).$$

share|cite|improve this answer

You can do the rotation as Moron says, or you can write $y=2x+\sin(x)$. This will remain a function, but doesn't have the same shape as a sine wave. It depends upon what you want.

share|cite|improve this answer
I've tried this and it yields okay results for what I want to do. I might use this for something else. I'd been trying to multiply when I should have been adding... – Omar Kooheji Jan 12 '11 at 17:45

Yes you can, but it might not be a function. Say y = f(x) is the curve you want to rotate. Then the equation of the curve of f(x) rotated by n radians is: ycos(n) - xsin(n) = f(ysin(n) + xcos(n)) Try it out here:

share|cite|improve this answer

protected by Zev Chonoles May 28 at 3:48

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.