Parametric equation of an ellipse How do I show that the parametric equations
$$x(t) = \sin(t+a)$$
$$y(t) = \sin(t+b)$$
define an ellipse?
I tried graphing it and I'm certain it is a rotated ellipse.
My first idea is to write it as
$$x(t) = \cos a \sin t + \sin a \cos t$$
$$y(t) = \cos b \sin t + \sin b \cos t$$
so the vector (x,y) is the vector (cos t, sin t) left multiplied by the matrix
$$\begin{pmatrix} \sin a & \cos a \\ \sin b & \cos b \end{pmatrix}$$
If I can show that this matrix is actually some enlargement followed by a rotation, I'm done, but I can't do this (I get contradictions).
 A: Here is how you advance
$$ x = \sin(t+a) \implies t=\sin^{-1}(x)-a . $$
Substituting in the other equation gives
$$ y = \sin( \sin^{-1}(x)+ b -a ) = \sin(\sin^{-1}(x))\cos(b-a)+\cos(\sin^{-1}(x))\sin(b-a) $$
$$ \implies y = x \cos(b-a) + \sqrt{1-x^2}\sin(b-a) $$
$$ \implies y-x \cos(b-a)=\sqrt{1-x^2}\sin(b-a) $$
$$ y^2+x^2-2\cos(b-a)\,xy-\sin^2(b-a)=0  $$
Now, Comparing with conic section general equation 
$$ Ax^2 + Bxy + Cy^2 +Dx + Ey + F = 0\text{ with }A, B, C\text{ not all zero,}$$
we have
$$ A = 1,\quad B = -2\cos(b-a),  \quad C = 1. $$
The equation represents an ellipse if $B^2-4AC<0$, that is
$$ 4\cos^2(b-a)-4= -4(1-\cos^2(b-a))= -4\sin^2(b-a)<0. $$
Note: 
$$ \cos(\sin^{-1}(x))= \sqrt{ 1- \sin(\sin^{-1}(x))^2 }=\sqrt{1-x^2}. $$
A: The parametrization represents an ellipse centered at the origin, albeit tilted with respect to the axes.  (You can demonstrate by plotting a few for yourself.)  The general form of this ellipse is
$$A x^2 + B x y + C y^2 = 1$$
The idea is to find the coefficients; this is done by expanding the sines and forming $x^2$, $y^2$, and $x y$.  I leave the algebra to the reader; I get
$$\left (\begin{array} \\ \cos^2{a} & \cos{a} \cos{b} & \cos^2{b} \\ \cos{2 a} & \cos{(a+b)} & \cos{2 b} \\\sin{2 a} & \sin{(a+b)} & \sin{2 b} \end{array}\right ) \cdot \left ( \begin{array}\\A\\B\\C \end{array} \right ) =  \left ( \begin{array}\\1\\0\\0 \end{array} \right )$$
One nontrivial point I should mention is that you get coefficients of $\sin^2{t}$, $\cos^2{t}$, and $\sin{t} \cos{t}$.  To get an expression that can be set to $1$ on the RHS, I took $\cos^2{t} = 1-\sin^2{t}$ and that did the trick.  You then get an expression independent of $t$ set to $1$, a coefficient of $\sin^2{t}$ set to zero, and a coefficient of $\sin{t} \cos{t}$, set to zero.
Inverting this matrix and performing the multiplication by the RHS, then rearranging the resulting equation for the ellipse, I get
$$x^2+y^2-2 \cos{(a-b)} x y = \sin^2{(a-b)}$$
as the sought ellipse.  Note that this is indeed an ellipse because $\cos^2{(a-b)} < 1$.
I also note that one finds parametrization like this in discussions of elliptical polarization of light.  Note that when $a-b = \pi/2$, the expression reduces to a circle; this is what physicists call circular polarization.
A: The matrix approach will be nicer but here is a brute force way.
$x=\sin(t+a)=\sin t \cos a+\cos t \sin a$
$y=\sin(t+b)=\sin t \cos b+\cos t \sin b$
solve for $\sin t$ and $\cos t$
$y\cos a  - x\cos b = \cos t (\sin b \cos a - \cos b \sin a) $
$y\sin a  - x\sin b = \sin t (\sin a \cos b - \cos a \sin b) $
Now square the last two and add to get
$x^2+y^2-2xy \cos(a-b)= \sin^2(a-b)$
Now you need to figure angle of rotation, etc.
