Circumsinusoidal Regions, Part One

circumsinusoidal region

The inner boundary of the yellow regions above is a sine curve (technically, a cosine curve, but that’s the same thing, just with a phase shift). The outer boundaries are semicircles. In order for this to work, to form these yellow regions, the semicircle centers (centers of the circles they are each half of) must be directly below peaks, and above troughs, of the sine (or cosine) curve, and vertically positioned at what would be called the rest position in physics. (I’m resorting to use of some physics terminology here, simply because I don’t know the corresponding mathematical terms).

In addition, each semicircle involved must have a radius equal to one-fourth the wavelength of the sine or cosine wave. The two sets of curves cross each other at the rest position, and are tangent to each other at each peak and trough, producing four of these yellow regions per wavelength.

In this case, semicircles could used because I adjusted the wavelength, making it exactly four times the amplitude of the wave. My goal was to compare the two curves, simply to see how well one simulates the other (answer:  not very well at all).  Then, however, I became more interested in the discrepancy between the two, represented by the yellow regions which are outside the true wave, and inside the semicircles which contain that wave. Until and unless I find that such regions already have a different name, I am naming these two-dimensional curved shapes “circumsinusoidal regions.” There are four of them per wavelength of the wave, and two per semicircle. Each circumsinusoidal region has two vertices, but the two paths connecting them are distinct curves. No part of either path contains any length which is a straight segment.

It would be possible to generate interesting solids by rotating circumsinusoidal regions around vertical or horizontal lines, such as the x- or y-axes, or around diagonal lines. Many such solids would be variations of a torus, including the central hole of a torus, but with circumsinusoidal cross-sections replacing a torus’s circular cross-sections. Unfortunately, I do not have the software I would need to generate pictures of such solid figures.

If the wavelength used for a given sinusoidal wave is not exactly four times the wave’s amplitude, semicircles won’t work to enclose the wave with the same points of tangency, but it is still possible to generate circumsinusoidal regions — using something, in their place, other than semicircles. This will be described in part two, which will be the next post on this blog.

A Half-Solved Mystery: Rotating a Sine Wave

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A Half-Solved Mystery

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.]

Basic Trigonometric Functions, Viewed On a Polar Coordinate System

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Basic Trigonometric Functions, Viewed On a Polar Coordinate System

The last post made me curious about other trigonometric functions’ graphs, in a polar coordinate system. They were not what I expected. Here they are.

The 27-63-90 Triangle

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The 27-63-90 Triangle

I’m not as pleased about finding this special right triangle as I was with the previous ones. For one thing, I didn’t use a regular polygon to derive it, but instead used the 36-54-90 triangle I had previously found, and applied a half-angle trigonometric identity to find the tangent of 27 degrees, as the tangent of half of 54 degrees. According to this identity, tan(27⁰) = (1 – cos(54⁰))/sin(54⁰). Once this gave me the leg lengths, I simply used the Pythagorean Theorem to determine the length of the hypotenuse. Finally, I checked all of this using decimal approximations.

The triple-nested radical in the expression for the hypotenuse is no cause for celebration, either. If anyone knows of a way to put this in simpler exact terms, please let me know.

The 15-75-90 Triangle

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The 15-75-90 Triangle

As the 30-60-90 triangle is based on an equilateral triangle, the 45-45-90 triangle is based on a square, the 18-72-90 and 36-54-90 triangles are based on the regular pentagon (see https://robertlovespi.wordpress.com/2013/03/12/the-18-72-90-and-36-54-90-triangles/), and the 22.5-67.5-90 triangle is based on the regular octagon (see previous post), so the 15-75-90 triangle is based on the regular dodecagon, shown here with three radii (red) and a single diagonal (purple). The 15-75-90 triangle is shown in yellow. An argument from symmetry is sufficient to show that angle EFC is the right triangle in this triangle, and the larger of its two acute angles (angle FCE) is one-half of an interior angle of this dodecagon. The interior angle of a regular decagon measures 150 degrees (the proof of this is trivial), and so angle FCE must measure half that amount, or 75 degrees. This leaves 15 degrees for angle CEF, via the triangle sum theorem.

What about the side lengths of the 15-75-90 triangle, though? First, consider the red diagonals shown, and let them each have a length of 2. Angles DAF and FAE each measure 30 degrees, since 360/12 = 30, and they are central angles between adjacent radii. This makes angle DAE 60 degrees by angle addition, and triangle DAE is known to be isosceles, since the two red sides are radii of the same regular dodecagon, and therefore are congruent. By the isosceles triangle theorem and triangle sum theorem, then, angles ADE and AED each also measure (180-60)/2 = 60 degrees, so triangle ADE is therefore equilateral, with the purple side, DE, also having a length of two. Symmetry is sufficient to see that DE is bisected by radius AC, which leads to the conclusion that EF, the long leg of the 15-75-90 triangle, has a length of 1.

Segment AF is a median, and therefore also an altitude, of equilateral triangle ADE, and splits it into two 30-60-90 triangles, one of which is triangle AEF. Its hypotenuse, AE, is already known to have a length of 2, while its short leg, EF, is already known to have a length of 1. Segment AF is therefore the long leg of this 30-60-90 triangle, with a length of √3.

AF, length √3, and FC, the short leg of the 15-75-90 triangle, together form dodecagon radius AC, already set at length 2. By length subtraction, then, FC, the 15-75-90 triangle’s short leg, has a length of 2 – √3. A test is prudent at this point, by taking the tangent of the 15 degree angle FEC in the yellow triangle. Tan(15 degrees) is equal to 0.26794919…, which is also the decimal approximation for FC/EF, or (2 – √3)/1.

All that remains to know the length ratios for the sides of the 15-75-90 triangle is to determine the length of EC, its hypotenuse, via the Pythagorean Theorem. The square of length EC must equal the square of 1 plus the square of (2 – √3), so EC, squared, equals 1 + 4 – 4√3 + 3, or 8 – 4√3. The hypotenuse (EC) must therefore be the square root of 8 – 4√3, which is √(8-4√3)) = 2√(2-√3)).

The short leg:long leg:hypotenuse ratio in a 15-75-90 triangle is, therefore, (2-√3):1:2√(2-√3)).

The 18-72-90 and 36-54-90 Triangles

It is well-known that an altitude splits an equilateral triangle into two 30-60-90 triangles, and that a diagonal splits a square into two 45-45-90 triangles. The properties of these “special right triangles,” as they are often called, are well-understood, and shall not be described here.

What happens if other polygons are split by diagonals, altitudes, or pieces thereof? Can more triangles be found which can allow, for example, exact determination of certain trigonometric ratios?

Yes, and the logical place to start looking is in the regular pentagon.

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In this diagram, the yellow triangle is the 18-72-90 triangle. Its hypotenuse is a diagonal of the pentagon, and its short leg is a half-side of the pentagon. Since sides and diagonals of regular pentagons are in the Golden Ratio, (1 + √5)/2, these two sides must be in twice that ratio. Let their lengths, then, be 1 (short leg) and 1 + √5 (hypotenuse), for those are simple, and in the specified ratio. The Pythagorean Theorem may then be applied to find the length of the long leg; the result is sqrt((2√5) + 5). Yes, nested radicals appear at this point, and they resist efforts to make them go away. No one promised this would be simple!

The blue triangle is the 36-54-90 triangle. Its long leg is a half-diagonal of the pentagon, while its hypotenuse is a full side of the pentagon. These triangle sides must, therefore, be in half the Golden Ratio, so the simplest lengths for those sides (which work) are 1 + √5 for the long leg, and 4 for the hypotenuse. Applying the Pythagorean Theorem to find the length of the short leg, nested radicals appear again in the solution:  sqrt(10 – 2√5).