Let us consider a cubic equation with real coefficients \begin{equation} y^3+py+q=0 , \label{eq:full} \end{equation}and let $R=q^2/4+p^3/27$ .

In the irreducible case when $R<0$ , Cardano's formulae give the solutions of this equation in terms of algebraic operations ( square and cube roots ) only apparently, because the argument of the cube roots is complex and we need goniometric functions to calculate it. The final solutions may be even rational, but we need the transcendental goniometric functions sine and cosine to calculate them.

Goniometric solutions are possible also in the case $R>0$ of just one real solution, as we shall see.

It was the French mathematician FRANÇOIS VIÈTE ( 1540 - 1603 ) in 1593, who first realized the analogy between equation () with $p<0$ and the triple angle formula $$ 4\cos^3 \frac{\vartheta}{3} -3 \cos \frac{\vartheta}{3} = \cos \vartheta . $$ Afterwards also ALBERT GIRARD ( 1632 ) in 1629 and RENÉ DESCARTES ( 1596 - 1650 ) in 1637 dealt with the solution of the irreducible case by goniometric functions.

Since the solutions in this case are already written in the article on Cardano's formulae, here we shall derive the goniometric solutions only for the case $R>0$ of just one real solution.

When $R=0$ or $p=0$ there is no difficulty, as explained in the simple analytic discussion of the cubic equation.

Therefore we shall assume that $R>0$ from now on and remember that the equation () has one real solution and two complex conjugate solutions in this case. We shall divide two cases.

First case. When $p<0$ , we shall set $$ \sqrt{-\frac{p^3}{27}}=\frac{q}{2}\sin\varphi , ~~~~~~\mathrm{with}~ -\frac{\pi}{2} < \varphi<\frac{\pi}{2} , $$ and note that $$ \sqrt{-\frac{p}{3}}=\sqrt[3]{\frac{q}{2}\sin\varphi} . $$

In Cardano's formulae this yields \begin{eqnarray*} \sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}} &=& \sqrt[3]{-\frac{q}{2}+\frac{q}{2}\cos\varphi}= -\sqrt{-\frac{p}{3}}\sqrt[3]{\frac{1-\cos\varphi}{\sin\varphi}} = -\sqrt{-\frac{p}{3}}\sqrt[3]{\tan\frac{\varphi}{2}} , \\ \sqrt[3]{-\frac{q}{2}-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}} &=& \sqrt[3]{-\frac{q}{2}-\frac{q}{2}\cos\varphi}= -\sqrt{-\frac{p}{3}}\sqrt[3]{\frac{1+\cos\varphi}{\sin\varphi}} = -\sqrt{-\frac{p}{3}}\sqrt[3]{\cot\frac{\varphi}{2}} , \end{eqnarray*}where we used the half angle formulae in the last step.

Now we shall introduce a new angle $\psi$ , $-\pi/2 < \psi<\pi/2$ , so that $$ \tan \psi = \sqrt[3]{\tan\frac{\varphi}{2}} . $$

For the real root we get $$ y_1 = -\sqrt{-\frac{p}{3}}\,(\tan\psi+\cot\psi)=-\frac{2\sqrt{-p/3}}{\sin(2\psi)} , $$ and for the two complex conjugate roots, we get $$ y_{2,\,3} = \sqrt{-\frac{p}{3}}\,\left[ \frac{1}{\sin(2\psi)} \pm i \sqrt{3}\cot(2\psi) \right] . $$

Second case. When $p>0$ , $p^3/27$ may be either greater or smaller than $q^2/4$ . Then we shall introduce an angle $\varphi$ , $0 < \varphi<\pi/2$ , so that $$ \sqrt{\frac{p^3}{27}}=\frac{q}{2}\tan\varphi . $$

In Cardano's formulae this yields \begin{eqnarray*} \sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}} &=& \sqrt[3]{\frac{q}{2}\frac{1-\cos\varphi}{\cos\varphi}}= \sqrt[3]{\frac{q\sin^2(\varphi/2)}{\cos\varphi}} = \sqrt{\frac{p}{3}}\sqrt[3]{\tan\frac{\varphi}{2}} , \\ \sqrt[3]{-\frac{q}{2}-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}} &=& \sqrt[3]{-\frac{q}{2}\frac{1+\cos\varphi}{\cos\varphi}}= -\sqrt[3]{\frac{q\cos^2(\varphi/2)}{\cos\varphi}} = -\sqrt{\frac{p}{3}}\sqrt[3]{\cot\frac{\varphi}{2}} . \end{eqnarray*} Now we shall introduce a new angle $\psi$ , $0 < \psi<\pi/2$ , so that $$ \tan \psi = \sqrt[3]{\tan\frac{\varphi}{2}} . $$

For the real root we get $$ y_1 = \sqrt{\frac{p}{3}}\,(\tan\psi-\cot\psi)=-2\sqrt{\frac{p}{3}}\cot(2\psi) , $$ and for the two complex conjugate roots, we get $$ y_{2,\,3} = \sqrt{\frac{p}{3}}\,\left[ \cot(2\psi) \pm i \frac{\sqrt{3}}{\sin(2\psi)} \right] . $$

Bibliography

1

Smirnov, Vladimir, 1972. Cours de mathématiques supérieures. Tome I. 2nd ed. Moscow: Éditions MIR.