Consider a standard sine curve.

It is periodic with a period of 2pi, and each period has two, for lack 
of a better term, lobes, one concave down, one concave up. Revolving a 
lobe about the x-axis results in a somewhat lemon-shaped solid, and 
revolving the whole sine curve about the x-axis forms a beaded strand of 
these solids. Alternatively, revolving a lobe about a vertical line at a 
odd multiple of pi/2 results in a bowl shape.

let's consider the lobe from x = 0 to x = pi.

The derivative of sin(x) is cos(x), and evaluating cos(x) at x = 0 and x 
= pi, we find the tangents of sin(x) at those values to have slopes of 
+1 and -1 respectively. This implies that the tangents intersect in a 
right angle, and that four lobes could be joined endpoint-to-endpoint to 
form a smooth, convex closed curve.

This curve would have the symmetry of a square, and like a square, could 
form two different solids of revolution, one analogous to a bicone, the 
other analogous to a cylinder.

But perhaps the more interesting solids are ones where three of these 
curves, placed in mutually orthogonal planes form a skeleton of sorts. 
this can be done in two ways that result in octahedral symmetry: the 
curves intersect where the lobes are joined, or they intersect at the 
crests of the lobes... comparing to squares again, this is analogous to 
either three orthogonal squares intersecting along the lines through 
opposite edge midpoints or intersecting along their diagonals.

One could simply take the convex hull of these skeletal forms, but I 
think the more interesting construction would be to create a stack of 
similar curves atop one of the curves where the joints or the crests 
trace the other two curves... similar to building up a sphere from three 
orthogonal circles and stacking smaller circles atop one circle while 
following the circumferences of the other two... though, would the 
resulting solid have octahedral symmetry still?

And can the amplitude and wavelength of a sine curve be manipulated such 
that lobes can be smoothly joined to form curves analogous to other 
regular polygons or even star polygons?

I've titled this musing "sinoids", and while sinoid doesn't appear to be 
defined in the mathematical literature, I'm not sure if it's the plane 
curves or the solids that the term would better fit.

Addendum 2024/08/25:

Let's consider a lobe of the standard sine curve positioned so it's
 endpoints are on the x and y-axes and it's crest is on the line y=x.

The chord of thelobe has length pi and forms the hypotenuse of a
45-45-90 right triangle with the x and y-axes. As such, the x and
y-intercepts are at (pi*sqrt(2))/2. Furthermore, the midpoint of the
curve is pi/2 from the origin along the line y=x, and the crest is at
pi/2+1.

Such a lobe is comparable to a quarter circle centered at the origin,
and four such lobes are comparable to a circle, though as 

pi/2 + 1 > (pi*sqrt(2))/2

The radius isn't constant and is greater at the cusps of a lobe than at
the points where lobes are joined.

As such, let's consider approximating this lobe with a quadrant of a
supercircle, that is, the curve:

x^n + y^n = r^n

where n=2 gives us the classic circle.

If we consider the supercircle n = 3 and r = 1, then the crest of the
quadrant would go though the point x=y=cube root of a half and it's
distance from the origin would be the square root of twice the square of
this cube root. Representing the cube root as a rational power, this is
written symbolically as

sqrt(2*(0.5^1/3)^2)


replacing 3 with x and setting the crest's distance from the origin to
the value listed above divided by its intercepts, for the sine lobe, we
get:

sqrt(2*(0.5^1/n)^2) = sqrt(2)/2 + sqrt(2)/pi

squaring both sides

2*(0.5^1/n)^2 = 1/2 +2/pi + 2/pi^2

divide both sides by 2

(0.5^1/n)^2 = 1/pi^2 + 1/pi + 1/ 1/4

Combine fractions on the right

(0.5^1/n)^2 = (4 + 4pi + pi^2)/4pi^2

square root both sides:

0.5^1/n = (pi + 2)/2pi

I'm not sure how to simplify this further(I think logarithms would be
involved, but I've forgotten what I've learned about manipulating
equations with logs), but rearranging the equation to make it equal 0
and plugging various values in for n into Google's calculator, I've
determined n is between 3.45684 and 3.45685... subtracting the nth root
of a half from the right side gets about 4e-7 for the last digit of n
being 4 and -6E-8 for the last digit being 5, which is about as much
precision as I care for a numerical approximation in the absence of an
exact formula. Also, I have no idea how to tell how well this
supercircle would approximate the sine lobe anywhere other than its
crest and endpoints.

Also, the crest of said lobe would have coordinates according to the following equation:

(pi/2 + 1)^2 = 2x^2


square root both sides:

pi/2 + 1 = sqrt(2)x

divide both sides by sqrt(2):

(pi/2 + 1)/sqrt(2) = x

put in proper form

x = (pi*sqrt(2) + sqrt(2))/2 ~ 2.9285

with y being the same.