## A Half-Solved Mystery: Rotating a Sine Wave

A few minutes ago, I wondered how to write a function whose graph would be a sine curve, but one that undulated above and below the diagonal line y=x, rather than the x-axis, as is usually the case. How to accomplish such a 45 degree counterclockwise rotation?

Well, first, I abandoned degrees, set Geometer’s Sketchpad to radians, and then simply constructed plots for both y = x and y = sin(x). Next, I added them together. The result is the green curve (and equation) you see above.

This only half-solves the problem. Does it undulate above and below y=x? Yes, it does. However, if you rotate this whole thing, clockwise, one-eighth of a complete turn, so that you are looking at the green curve going along the x-axis, you’ll notice that it is not a true sine curve, but a distorted one. Why? Because it was generated by adding y-values along the original x-axis, not by a true rotation.

I’m not certain how to correct for this distortion, or otherwise solve the problem. If anyone has a suggestion, please leave it in a comment. [Note: an astute follower of this blog has now done exactly that, so I refer the reader to the comments for the rest of the story here.]

I go by RobertLovesPi on-line, and am interested in many things, the majority of which are geometrical. Welcome to my little slice of the Internet. The viewpoints and opinions expressed on this website are my own. They should not be confused with the views of my employer, nor any other organization, nor institution, of any kind.
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### 8 Responses to A Half-Solved Mystery: Rotating a Sine Wave

1. I would define x’ = x cos(pi / 4) + y sin(pi / 4) and y’ = x sin(pi / 4) – y cos (pi / 4). This rotates the axes 45 degrees from (x,y) to (x’,y’).

Then I would graph y’ as a function of x’. I tried it on Excel just to be sure. I made a column of x varying from 0 to 6.3 in steps of 0.1, a column of y defined as sin(x), and two columns for x’ and y’, then graphed y’ as a function of x’ in an xy scatter plot.

(It also simplifies to x’ = (x + y) / sqrt(2) and y’ = (x – y) / sqrt(2).)

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• Thanks for the help! I’ll update the post to redirect people to the comments for, well, the end of the story.

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• Thank you for stirring my interest. 🙂

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• Anonymous says:

Hi – great post. Frankly, I still do not understand how this corrects your issue. So instead of “h(x) = sin(x) + x” what should the function be?

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2. Anonymous says:

do you know how to make the diagonal go where ever you want like have it more vertical or horizontal

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• Well, all I did was add y = x to the sine function. For other slopes, use y = kx instead of y = x, with k not equal to one. This still only “half-solves” it, though; the resultant wave is skewed.

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3. Anonymous says:

I was able to find an implicit function that graphs what you want. y=sqrt(2)(sin((x+y)/sqrt(2))+x

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