I am trying to convert the parametric function

$x(t) = a\cdot(t - \sin(t)) + b\cdot\cos\left(\frac{t}{2}\right)$

$y(t) = a\cdot\cos(t) + b\cdot\sin\left(\frac{t}{2}\right)$

into a cartesian form. As far as I understood, the usual approach is to solve one equation for t and insert it into the second one.

Wolfram Alpha gives an absurdly complex solution when trying to solve y(t) for t:

Solution of Wolfram Alpha

Is this really a valid and exact solution? It looks a bit like an approximation to me. And is there indeed no simpler form?

Perhaps the whole term could be simplified a lot after inserting it into x(t), but the terms grow much too large for me to try it.


If I let Wolfram Alpha solve the term

$y(t) = a\cdot\cos(t)+b\cdot\sin\left(\frac{t}{2}\right)+c$

the result is much simpler: Solution 2 of Wolfram Alpha

What am I missing here?

  • $\begingroup$ It is very complex and the result is OK. $\endgroup$ – Claude Leibovici Jun 9 '16 at 8:50
  • $\begingroup$ @ClaudeLeibovici I tried with a slightly modified version and got a much simpler solution, see my edit above. Can you imagine why this is the case? $\endgroup$ – Ctx Jun 9 '16 at 10:03

How this can be made : consider $$y = a\cos(t)+b\sin\left(\frac{t}{2}\right)+c$$ and rewrite it as $$y=a\left(1-2\sin^2\left(\frac{t}{2}\right)\right)+b\sin\left(\frac{t}{2}\right)+c$$ Now, define $z=\sin\left(\frac{t}{2}\right)$ to get $$y=a(1-2z^2)+bz+c$$ $$2 a z^2-b z-(a+c-y)=0$$ Solve the quadratic. Then $t=2\sin^{-1}(z)$.

Now, why does Wolfram Alpha reacts so differently ? Just ask them !

Have fun !

  • $\begingroup$ That helped, thank you very much $\endgroup$ – Ctx Jun 9 '16 at 11:58

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