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:
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.
Edit:
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?
Best Answer
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 !