# Using a rotated function directly

I want to be able to directly use the resulting function of rotating a sine function.

My original function is:

f(x) = B * sin(x)

My domain for x is -pi to pi.

My domain for B is -1 to 1.

When I rotate it by theta degrees I have (actually my theta is fixed and is 45 degrees):

x' = cos(theta) * x - sin(theta) * B * sin(x) y' = sin(theta) * x + cos(theta) * B * cos(x)

The problem is now I have a new function depending on x but that has at its source x' and not x anymore. That's not what I need.

If I try to substitute x in the y' function I end up not being able to use my previous x because I can not solve it in a closed form.

Am I missing something here? Is there any mathematical trick to do this job? I have already tried parametric equations to no avail.

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The rotated version of the sine curve is no longer necessarily a function; it no longer passes the vertical line test in general... –  Guess who it is. Nov 7 '11 at 13:48
Yes you're right, if my angle were bigger than 45 and my domains were different. That's not the case. –  hildebrando Nov 7 '11 at 14:45

## 1 Answer

It looks like you are mixing up the rotation and the function. You have $y=B\sin x$ as your function. Now you want to rotate to coordinates $(x',y')$ with a rotation $x'=x \cos \theta - y \sin \theta, y'=y \cos \theta + x \sin \theta$. You can certainly write your function in the new coordinates (and as long as $|\theta| \le 45^{\circ}$ it will still be a function). If you want to get rid of the unprimed variables, you need the equations that go in the opposite direction: $x=x' \cos \theta + y' \sin \theta, y=y' \cos \theta - x' \sin \theta$. You can insert these, giving $y' \cos \theta - x' \sin \theta=B \sin (x' \cos \theta + y' \sin \theta)$, expand the $\sin$ on the right and collect terms.

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If I understood you correctly after expanding the sin and collecting terms I end up with another non-solvable function for y'. –  hildebrando Nov 7 '11 at 16:14
You have a transcendental equation for $x$, i.e., one involving $x$ and $\sin x$ or $\cos x$. In general, there are no closed form solutions to transcendental equations. –  Jeff Nov 8 '11 at 21:16