On 2014-05-14, Ailurus wrote:
> By "nontrivial way", do you mean the cosine or the square root (or both)?
I didn't actually notice sqrt.
Such equations are in fact not called quadratic, that was confusing.
Your original equations also contained i, which is, I presume,
sqrt(-1)?
> And yes, I cou
By "nontrivial way", do you mean the cosine or the square root (or both)?
And yes, I could eliminate b and then solve for a, but that's a manual step
I'd rather avoid.
Ok, so I simplified the expressions using Sage (I'm quite surprised to see
that simplify_full() produces better results than M
On 2014-05-14, Ailurus wrote:
> Hi all,
>
> I'm trying to solve the following system of two equations,
>
> Eq1 = 2*a^2*cos(pi/n)^2 - 2*a - b - (a*sin((2*pi)/n)*(-2*(cos((2*pi)/n) -
> sin((2*pi)/n)*i)*(a^2*cos((2*pi)/n) - 4*a - 2*b + a^2 + 2))^(1/2))/2 +
> 2^(1/2)*a*cos(pi/n)^2*(-(cos((2*pi)/n) -
