How to solve a quartic equation using trigonometric power identitiy?

Question

According to this post, we can solve a cubic equation

$$t^3+pt+q=0$$

by the trigonometric identity

$$4\cos^3\theta-3\cos\theta-\cos3\theta=0$$

So I've tried to solve the quartic equation

$$t^4+pt^2+qt+r=0$$

using the identity

$$8\cos^4\theta-4\cos2\theta-\cos4\theta-3=0$$

$$8\cos^4\theta-4(2\cos^2\theta-1)-\cos3\theta\cos\theta+\sin3\theta\sin\theta-3=0$$

$$8\cos^4\theta-8\cos^2\theta-\cos3\theta\cos\theta+1+\sin3\theta\sin\theta=0$$

If we let $t:=A\cos\theta$, then we have

$$A^4\cos^4\theta+A^2p\cos^2\theta+Aq\cos\theta+r=0$$ or $$8\cos^4\theta+\frac{8p}{A^2}\cos^2\theta+\frac{8q}{A^3}\cos\theta+\frac{8r}{A^4}=0$$

now $\frac{8p}{A^2}=-8$ implies that $A=\sqrt{-p}$ and we need to find $\theta$ in the following system

$$\begin{cases}\cos3\theta=-\frac{8q}{A^3}\\1+\sin3\theta\sin\theta=\frac{8r}{A^4}\end{cases}$$

What would you think on solving this system? Is it possible to solve this system? I couldn't find anything on the web about the solving a quartic equation by trigonometric power identity, so I don't know if this way works ...

Answer 1

Four-angles formula:

$\cos{4\theta}=1-8\cos^2\theta+8\cos^4\theta$.

Quartic equation:

$p = q x + r x^2 + s x^3 + t x^4$

Let the cubic resolvent is

$p^3 (s^3 - 4 r s t + 8 q t^2) = p^2 t (-q s^2 + 4 q r t + 8 p s t)V + p t^2 (q^2 s + 4 p r s - 8 p q t)V^2 + t^3 (q^3 + 4 p q r + 8 p^2 s)V^3$

Trigonometric solution:

$U = 1/(2 p^2 (3 s^2 - 8 r t) + 4 p t (q s + 16 p t) V + 2 t^2 (3 q^2 + 8 p r) V^2)$

$L_0 = p^4 (3 s^4 - 16 r s^2 t + 64 q s t^2 + 256 p t^3)$

$L_1 = 4 p^3 t (q s^3 - 8 q r s t - 32 p s^2 t + 48 q^2 t^2 + 128 p r t^2)$

$L_2 = 2 p^2 t^2 (-7 q^2 s^2 - 24 p r s^2 + 24 q^2 r t + 128 p r^2 t - 128 p q s t - 256 p^2 t^2)$

$L_3 = 4 p t^3 (q^3 s + 8 p q r s + 48 p^2 s^2 - 32 p q^2 t - 128 p^2 r t)$

$L_4 = t^4 (3 q^4 + 16 p q^2 r + 64 p^2 q s + 256 p^3 t)$

$W = L_0 + L_1 V + L_2 V^2 + L_3 V^3 + L_4 V^4$

$S_2 = \cos{((\arccos{(1 + 8 W U^2)} + 2 \pi j)/4)}$

$j=0,1,2,3$

$S = (-p s + q t V) U^{1/2}$

$T = p t U^{2/3}$

$z = (S_2 - S)/(4 T)$

$M = z U^{1/6}$

$x = (-M \pm (M^2 - 4 V)^{1/2})/2$

But this solution gives eight roots, four true and four excess.

Wolfram-code for verifing:

{{p, q, r, s, t} = RandomInteger[{-1000, 1000}, 5];
  Eq = -p + q x + r x^2 + s x^3 + t x^4;
  Print["\nEquation: 0 = ", Eq, "\n\nSolution by CAS:"];
  Print[x /. (Eq // NSolve) // Sort, "\n"];
  Print["Solution by formula:"];
  {f, a, b, c} = {p^3 (s^3 - 4 r s t + 8 q t^2), 
    p^2 t (-q s^2 + 4 q r t + 8 p s t), 
    p t^2 (q^2 s + 4 p r s - 8 p q t), t^3 (q^3 + 4 p q r + 8 p^2 s)};
  {k, m, n} = {a b + 9 c f, b^2 - 3 a c, a^2 + 3 b f};
  If[m == 0 && n == 0, 
   Print["Formula does not work with double/triple roots"]; Goto[end]];
  If[m == 0, V = (-k + Sqrt[k^2 - 4 m n])/(2 n); S = 1; T = V];
  If[n == 0 || (m != 0 && n != 0), V = (-k + Sqrt[k^2 - 4 m n])/(2 m);
    S = V; T = 1];
  F = f - a - b - c;
  A = (a + 2 b + 3 c) S + (2 a + b - 3 f) T;
  B = (b + 3 c) S^2 + 2 (a + b) S T + (a - 3 f) T^2;
  Q = c S^3 + b S^2 T + a S T^2 - f T^3;
  If[Q == 0, Print["Formula does not work with double/triple roots"]; 
   Goto[end]];
  G = -2 B^3 + 9 A B Q + 27 Q^2 F;
  H = {G^(1/3), -(-1)^(1/3) G^(1/3), (-1)^(2/3) G^(1/3)};
  y = (H[[1]] - B)/(3 Q); V = (1 + S y)/(1 + T y);
  U = 1/(2 p^2 (3 s^2 - 8 r t) + 4 p t (q s + 16 p t) V + 2 t^2 (3 q^2 + 8 p r) V^2);
  L0 = p^4 (3 s^4 - 16 r s^2 t + 64 q s t^2 + 256 p t^3);
  L1 = 4 p^3 t (q s^3 - 8 q r s t - 32 p s^2 t + 48 q^2 t^2 + 128 p r t^2);
  L2 = 2 p^2 t^2 (-7 q^2 s^2 - 24 p r s^2 + 24 q^2 r t + 128 p r^2 t - 128 p q s t - 256 p^2 t^2);
  L3 = 4 p t^3 (q^3 s + 8 p q r s + 48 p^2 s^2 - 32 p q^2 t - 128 p^2 r t);
  L4 = t^4 (3 q^4 + 16 p q^2 r + 64 p^2 q s + 256 p^3 t);
  W = L0 + L1 V + L2 V^2 + L3 V^3 + L4 V^4;
  S2 = Table[Cos[1/4 (ArcCos[1 + 8 W U^2] + 2 \[Pi] j)], {j, 0, 3}];
  S = (-p s + q t V) U^(1/2); T = p t U^(2/3);
  z = (S2 - S)/(4 T); M = z U^(1/6);
  X = {(-M - (M^2 - 4 V)^(1/2))/2, (-M + (M^2 - 4 V)^(1/2))/2};
  (*Print[X//N];*)
  Do[If[Re[N[X[[i, 1]] + X[[j, 2]] + X[[k, 3]] + X[[l, 4]], 10]] == N[-s/t, 10],
    Print[{X[[i, 1]], X[[j, 2]], X[[k, 3]], X[[l, 4]]} // N // Sort]; Break[]
    ], {i, 2}, {j, 2}, {k, 2}, {l, 2}];
  Label[end];
  };

Answer 2

it`s possible to solve this system. Just rewrite $$\cos 3\phi = \cos\phi(2\cos2\phi-1)$$ and $$\sin3\phi \sin\phi=\sin^2(\phi)(2\cos 2\phi+1)$$

Answer 3

In general, you cannot parameterize a two-dimensional set, like $$ \left(-\frac{8q}{A^3},\frac{8r}{A^4}\right) $$ with a one-edimensional curve, like $$ (\cos3\theta,1+\sin3\theta\sin\theta) $$

Answer 4

This equation can be solved by trigonometric methods based on a certain geometric interpretation which does not include the graph of the quartic function (only as a decorative element), but there are still some ambiguities regarding the selection criteria for both the correct trigonometric expressions and the signs in the final solutions. Here I give an example that is suitable for equations that have two positive and two negative solutions. He also noted that the order of matching the solutions of the cubic equation in the relationships that give the values ​​of the angles $\displaystyle \omega_n$ is important.

$\displaystyle p=-2153, q=-13500, r=510796$

$\displaystyle R=\sqrt{\frac{q^2}{4r}-p}$

Resolvent cubic:

$\displaystyle y^3-py^2-4ry+4rp-q^2=0$

$\displaystyle y_1=-2072, y_2=-1528, y_3=1447$

You may notice that the resolvent does not know what the sign of the coefficient $\displaystyle q$ is. This means that in order to obtain the correct signs, the solutions of the quartic equation shown below have already been multiplied by $\displaystyle q/|q|$, but these multiplications do not appear there for reasons of simplicity.

Solution:

$\displaystyle \cos \omega_n=\sqrt{\frac{1}{2}+\sqrt{\frac{1}{4}-\frac{r}{y_n^2}}}$

$\displaystyle \sin \omega_n=\sqrt{\frac{1}{2}-\sqrt{\frac{1}{4}-\frac{r}{y_n^2}}}$

for $\displaystyle n=1, 2, 3$

$\displaystyle t_1=+2R \sin \omega_1 \sin \omega_2 \sin \omega_3 =+13$

$\displaystyle t_2=-2R \sin \omega_1 \cos \omega_2 \cos \omega_3 =-22$

$\displaystyle t_3=-2R \cos \omega_1 \sin \omega_2 \cos \omega_3 =-38$

$\displaystyle t_4=+2R \cos \omega_1 \cos \omega_2 \sin \omega_3 =+47$

Answer 5

The second way to get solutions in trigonometric form (without using trigonometric power indentity). Represent root's of quadratic equation $$ax^2+bx+c=0$$ in trigonometric form $$x=cos\left(\frac12 arccos\left(\frac {b^2-4ac-2a^2}{2a^2}\right)+\frac {\pi n} 2 \right)-\frac {b}{2a}$$ where $n=0,1$. Then rewrite $$t^4+pt^2+qt+r=0$$ in form of $$\left(t^2+\alpha t+\beta \right)\left(t^2-\alpha t+\frac {r}{\beta} \right)$$ Now, we can find $\alpha$, $\beta$, solving system of equations $$\begin{cases} \alpha^2(\alpha^2+a)^2-4c\alpha-b^2=0\\ \beta^2+\frac {b}{\alpha} \beta-c=0 \end{cases}$$