If I give you the curve $\gamma (t) = ( \cos (t), \cos(t+a))$, how can I obtain an implicit equation? or which change of basis can I do to get the canonical implicit form?
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I would try expressing $\sin t$ as a linear combination of $\cos t$ and $\cos(t+a)$, then substituting into $\cos^2 t+\sin^2 t=1$. – Aug 18 '18 at 01:58
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Please refer to this. – Ng Chung Tak Aug 18 '18 at 08:50
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Basically the same method as described in this answer: Expand $y=\cos(t+\alpha) = \cos t\cos\alpha-\sin t\sin\alpha$ and solve for $\sin t$ (disregard possible zero denominators for now). Substitute $x=\cos t$ into the result, then substitute both into $\cos^2t+\sin^2t=1$, rearrange and simplify.
amd
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This is related to MSE question 2881242. Define variables $ u := \cos(t), \, v := \cos(t+a), \, w := \cos(a). \,$ The implicit equation is $\, u^2 + v^2 + w^2 - 2 u v w = 1. $
Somos
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The canonical form is $A x^2 + B y^2 +2 C x y = 1$
Substitute $x=\cos(t)$ and $y = \cos(a)\cos(t)-\sin(a)\sin(t)$ into the above and try to eliminate all coefficients of $\sin(t)$ and $\cos(t)$.
It helps to use the substitutions $\cos^2(t) = \frac{\cos(2 t)+1}{2}$, $\sin^2(t) = \frac{1-\cos(2 t)}{2}$ and $\sin(t)\cos(t) = \frac{\sin(2 t)}{2}$
You will get in the end the following
$$ \begin{multline} \left( B \cos^2(a) + C \cos(a) + \frac{A-B}{2} \right) \cos(2 t) - \Bigl( B \sin(a)\cos(a)+C \sin(a) \Bigr) \sin(2 t)+ \\ + \left( C \cos(a) + \frac{A+B}{2} \right) = 1 \end{multline} $$
Now we eliminate all coefficients of $\sin(2 t)$ and $\cos(2t)$ to convert the above into a system of equations
$$\begin{aligned} B \cos^2(a) + C \cos(a) + \frac{A-B}{2} & = 0 \\ B \sin(a)\cos(a)+C \sin(a) & = 0 \\ C \cos(a) + \frac{A+B}{2} & = 1 \end{aligned} $$
Solve the above for $A$, $B$ and $C$.
John Alexiou
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