On my grapher, $(\cos t, \sin (t+1))$ generates a geometric figure. What is that figure?

Question

On Wolfram Alpha I am getting a graph like this: http://www.wolframalpha.com/input/?i=%28sin+t%2Ccos%28t%2B1%29%29

Is this an ellipse? I really don't know how to find it.

Answer 1

Yes it is : you can write $\sin(1+t)=\sin(1)\cos(t)+\sin(t)\cos(1).$ Then your graph is the set $$\{\begin{pmatrix}1&0\\\sin(1)&\cos(1)\end{pmatrix}\times\begin{pmatrix}\cos(t)\\\sin(t)\end{pmatrix} \mid t\in\mathbb{R}\} $$ which is a regular linear transformation of the unit circle.

Answer 2

Well in your WA submission you've got the functions switched around, but that's equivalent to a phase change so it doesn't really make much of a difference.

We're graphing the pair of functions $x = \cos{t}$ and $y = \sin{(t+1)}$, but let's expand and rearrange the second one using the trigonometric identity $\sin{(\alpha + \beta)} = \sin{\alpha}\cos{\beta} + \cos{\alpha}\sin{\beta}$. So:

\begin{eqnarray} y & = & \sin{(t+1)} \\ & = & \sin{t}\cos{1} + \cos{t}\sin{1} \\ y - \cos{t}\sin{1} & = & \sin{t}\cos{1} \\ y - x \sin{1} & = & \sin{t}\cos{1} \\ (y - x \sin{1})^2 & = & (\sin{t}\cos{1})^2 \\ & = & \sin^2{t}\cos^2{1} \\ & = & (1 - \cos^2{x})\cos^2{1} \\ & = & (1 - x^2)\cos^2{1} \\ y^2 - 2xy\sin{1} + x^2 \sin^2{1} & = & (1 - x^2)\cos^2{1} \\ x^2 + y^2 - 2xy\sin{1} & = & \cos^2{1} \end{eqnarray}

Which is, indeed, an ellipse, and Wolfram Alpha even identifies it as such.