Proving that x=f*sin(wt+θ), y=g*sin(wt+φ) traces an ellipse

[DarkEternal]

Ack, such a simple question, but I haven't worked with conic sections in years. Can anyone suggest an elegant way to show that

$$x=f*Sin(wt+\theta)$$
$$y=g*Sin(wt+\phi)$$

is an ellipse? I've tried using a rotation matrix on standard parametric ellipse equations and then solving for the angle of rotation and the axes sizes in terms of the variables but it seems messy. Then I tried getting it to fit the general equation but I'm not sure how that would work. However, a simpler method is eluding me. Any help?

 

[krab]

Some hints: This is not an upright ellipse like $x^2/a^2+z^2/b^2=1$, unless $\phi-\theta=\pi/2$. But you can get it into that form by applying a transformation like z=y+kx, where k is a constant. By expanding the sines, find k in terms f, g, $\phi$ and $\theta$ (actually will depend only on $\phi-\theta$ and f/g), and you are there.

 

[DarkEternal]

Can you clarify what you mean by a transformation of the form z=y+kx? what is y?

 

[Tide]

Ack, such a simple question, but I haven't worked with conic sections in years. Can anyone suggest an elegant way to show that

$$x=f*Sin(wt+\theta)$$
$$y=g*Sin(wt+\phi)$$

is an ellipse? I've tried using a rotation matrix on standard parametric ellipse equations and then solving for the angle of rotation and the axes sizes in terms of the variables but it seems messy. Then I tried getting it to fit the general equation but I'm not sure how that would work. However, a simpler method is eluding me. Any help?

If you don't mind a little nonrigorous math you can try the following.

Rewriting your equations slightly:

$$x = A cos(\omega t + \alpha)$$
$$y = B cos(\omega t + \beta)$$

The second equation gives
$$\omega t = \cos ^{-1} \left(\frac {y}{B}\right) - \beta$$

Substitute into the first equation:
$$x = A \cos( \cos ^{-1} \frac {y}{B} + \alpha - \beta)$$

Use the addition formulas for cosine and the basic identities for the inverse trig function and the equation can be cast into the familiar elliptical form.

 

[DarkEternal]

By familiar elliptical form, do you mean general form? I can't seem to get rid of that square root term from the inverse trig identity...

 

[Tide]

Rearrange terms to get the radical on one side of the equation then square both sides. The "general form" is $ax^2+bxy+cy^2+d=0$ and, depending on the signs and values of the various coefficients, will produce an ellipse, hyperbola or parabola.

 

[DarkEternal]

I got it, but thanks for your replies.