From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: Sinusoidal rotation??
Date: 5 Aug 1996 01:08:04 GMT

In article <DvIvpG.2wME@pen.k12.va.us>,
Michael A. Stueben <mstueben@pen.k12.va.us> wrote:

>Objective is to create a general equation for a sine curve that
>is rotated off the x-axis.  The new axis is a specified line so
>that the amplitudes of the sine are perpendicular to the new
>axis.

I'm not sure how large you mean this rotation to be. The slope of
the curve  y=sin(x)  at the origin is  1, so a rotation through an
angle of  pi/4  will make the curve perpendicular to the horizontal
axis at the origin. Is that what you mean?

>Theory:For a sine centered on the axis y=x, the equation is
>y-x=sin(y+x).

I assume you mean the label to be "Conjecture:". 

This is almost correct.
If a point  (a,b)  is in the new graph, its rotation  (c,d)  backwards
by an angle  pi/2  will be in the old graph, so that  (c,d)  satisfies the
original equation  d = sin(c). (I assume that's what you meant the
original curve to be.) Well, the relationship between  (c,d)  and  (a,b)
is linear:  c = a*cos(-pi/2) - b*sin(-pi/2), and
            d = a*sin(-pi/2) + b*cos(-pi/2).
If we let  r = sqrt(2)/2 = 0.7071...  then this is simply
	c = r(a+b)
	d = r(-a+b)
so the statement that  (c,d)  is on the original curve is the statement
that
	r(b-a) = sin(r*(b+a))
If you like, you may write this as
	(rb) - (ra) = sin( (rb) + (ra) )
so that the point  (ra, rb)  is on precisely the curve you proposed.
That is, your curve is the same set of points as the rotation you
wanted after the latter have been scaled back by a factor of  r.

I tried to present this in a way which would allow you to express the
result of rotating any curve (defined implicitly) through any angle.

>	We have no software capable of verifying this theory.
Software? Feh!


dave