What is the equation describing sin(x) rotated by an angle?

Question

I'm talking about a curve that looks like this:

I've come up with these equations, which seem to describe it:

$$ \left\{\begin{aligned} x &= t \cos \alpha - \sin t \sin \alpha\\ y &= t \sin \alpha + \sin t \cos \alpha \end{aligned}\right. $$

I want this to be a function $y(x)$, so $\frac{\mathrm{d} x}{\mathrm{d} t} \neq 0$. This forces $\alpha \in [0,\frac{\pi}{4})$.
Here's the hard part, how do I combine these two eqations into one? I can't use $\sin^2 x + \cos^2 x = 1$, nor can I solve for $t$ to plug it into the second equation.

Answer 1

We have

$$y\mathrm{cos}(\alpha)-x\mathrm{sin}(\alpha)=(t\mathrm{sin}(\alpha)+\mathrm{sin}(t)\mathrm{cos}(\alpha))\mathrm{cos}(\alpha)-(t\mathrm{cos}(\alpha)-\mathrm{sin}(t)\mathrm{sin}(\alpha))\mathrm{sin}(\alpha)$$$$=\mathrm{sin}(t)\mathrm{cos}^2(\alpha)+\mathrm{sin}(t)\mathrm{sin}^2(\alpha)=\mathrm{sin}(t)$$

And $$y\mathrm{sin}(\alpha)+x\mathrm{cos}(\alpha)=(t\mathrm{sin}(\alpha)+\mathrm{sin}(t)\mathrm{cos}(\alpha))\mathrm{sin}(\alpha)+(t\mathrm{cos}(\alpha)-\mathrm{sin}(t)\mathrm{sin}(\alpha))\mathrm{cos}(\alpha)$$$$=t\mathrm{sin}^2(\alpha)+t\mathrm{cos}^2(\alpha)=t$$

This gives us $$\mathrm{sin}\left(y\mathrm{sin}(\alpha)+x\mathrm{cos}(\alpha)\right)=y\mathrm{cos}(\alpha)-x\mathrm{sin}(\alpha)$$

You are unlikely to be able to get a relation between $y$ and $x$ that is more explicit than this.

Answer 2

You can't achieve that because you would need to invert

$$x=t\cos\alpha-\sin t\sin \alpha$$ and get $t(x)$, but this is a transcendental equation that has no analytical solution.