The parametrisation (cos(t),sin(t+a)) almost always seems to describe a rotated ellipse. How can I prove it?

Question

I was wondering how a curve would look like if you have two hands of a clock moving at the same speed from different positions while you track the x of one hand and the y of the other hand over time. That's how I got (cos(t),sin(t+a)) where t is the time and a the angle between the hands.

It turned out to almost always describe an ellipse tilted by 45°. I have no idea why it is always tilted by that angle and how to show this parametrisation describes an ellipse.

My failed tries: 1. Rotating (cos(t),sin(t+a)) by 45° and showing that the rotated points suffice the $\frac{x^2}{u^2}+\frac{y^2}{v^2}=1$ equation.

2. Using the distance to focal points definition for (cos(t),sin(t+a))

Maybe it's not possible to show in general because of the special cases where (cos(t),sin(t+a)) describes a line. Please help!

I also failed showing it for a special a.

Sry for my bad english. I'm no native speaker

Answer 1

Eliminate $t$ by expanding $y=\sin(t+a)$ into $y=\cos t\sin a+\cos a\sin t$ and then solve for $\sin t$ to get $(y-x\sin a)\sec a$, so that $$x^2+(y-x\sin a)^2\sec^2a=1,$$ which can be rearranged into the more standard-looking $$x^2-2xy\sin a+y^2=\cos^2a. \tag{*}$$ For $\cos a\ne0$, this equation indeed describes an ellipse. Using any of the usual methods, such as orthogonal diagonalization, its principal axes can be found to be in the directions $(1,1)$ and $(-1,1)$, i.e., its major axis is rotated 45° either clockwise or counterclockwise from standard position. When $\sin a=0$, you have a circle.

When $\cos a=0$, the implicit Cartesian equation (*) becomes $(x\pm y)^2=0$, i.e., a double line, but the original parametric equation only produces values of $x$ and $y$ in the interval $[-1,1]$, so the ellipse collapses to a line segment in this case.

Answer 2

Starting from the equation $x^2 -2xy \sin a +y^2 = \cos^2 a$ we can assume a point (x,y) to have the coordinates (u,w) in a 45° rotated coordinate system. (u,w) in the rotated system has the coordinates $(\frac{u-w}{\sqrt 2},\frac{u+w}{\sqrt 2})$ in the old system, so that $x=\frac{u-w}{\sqrt 2}$ and $x=\frac{u+w}{\sqrt 2}$. Replacing x and y in the staring equation yields $u^2+w^2-\sin a (u^2-w^2)= \cos^2 a$ which can be simplified to

$$ \frac{u^2}{\cos^2 a \over 1-\sin a} + \frac{w^2}{\cos^2 a \over 1+\sin a} = 1$$ which is an ellipse for $\cos^2 a$ not equal to $0$.

Answer 3

$$ (\cos(t),\sin(t+a))=(\cos(t),\sin(t)\cos(a)+\cos(t)\sin(a)) $$ $$ \big\Updownarrow $$ $$ (y-\sin(a)x)^2+\cos^2(a)x^2=\cos^2(a) $$ $$ \big\Updownarrow $$ $$ \begin{align} 2\cos^2(a) &=2\begin{bmatrix}x&y\end{bmatrix} \begin{bmatrix} 1&-\sin(a)\\ -\sin(a)&1 \end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix}\\ &=\begin{bmatrix}x&y\end{bmatrix} \begin{bmatrix} 1&1\\ 1&-1 \end{bmatrix} \begin{bmatrix} 1-\sin(a)&0\\ 0&1+\sin(a) \end{bmatrix} \begin{bmatrix} 1&1\\ 1&-1 \end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix}\\[6pt] &=(1-\sin(a))(x+y)^2+(1+\sin(a))(x-y)^2 \end{align} $$ Therefore, $$ \frac{\left(\frac{x+y}{\sqrt2}\right)^2}{1+\sin(a)}+\frac{\left(\frac{x-y}{\sqrt2}\right)^2}{1-\sin(a)}=1 $$ which is an ellipse with semi-major and semi-minor axes of $\sqrt{1+\sin(a)}$ and $\sqrt{1-\sin(a)}$ rotated by $\frac\pi4$.