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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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    Yes you're right, if my angle were bigger than 45 and my domains were different. That's not the case.2011-11-07

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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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    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.2011-11-08