# trigonometric cubic formula

Given a cubic polynomial of the form $f(X)={X}^{3}+a{X}^{2}+bX+c=0$, one may reduce $f(x)$ via the substitution $X\to (x-a/3)$ to obtain $\stackrel{~}{f}(x)=f(x-a/3)$ where the reduced polynomial^{} may be represented

$$\stackrel{~}{f}(x)={x}^{3}+qx+r$$ | (1) |

The roots to (1) are given by Viéte in the following cases:

Case I The roots of $\stackrel{~}{f}(x)$ are real:

Define $t\equiv \sqrt{-4q/3}$ and $\alpha =\mathrm{arccos}(-4r/{t}^{3})$.
Then the roots of $\stackrel{~}{f}(x)$ are

$$t\mathrm{cos}(\alpha /3),t\mathrm{cos}(\alpha /3+2\pi /3),t\mathrm{cos}(\alpha /3+4\pi /3)$$ |

Case II The roots of $\stackrel{~}{f}(x)$ are complex:

Keeping the definition of $t$ from Case I, if $-4q/3\ge 0$, then the real root of $\stackrel{~}{f}(x)$ is

$$t\mathrm{cosh}(\beta /3)\mathit{\hspace{1em}}\text{where}\mathit{\hspace{1em}}\mathrm{cosh}(\beta )=(-4r/{t}^{3})$$ |

If $$, then the real root of $\stackrel{~}{f}(x)$ is

$$t\mathrm{sinh}(\gamma /3)\mathit{\hspace{1em}}\text{where}\mathit{\hspace{1em}}\mathrm{sinh}(\gamma )=(-4r/{t}^{3})$$ |

One may then inverse^{} transform the roots of $\stackrel{~}{f}(x)$ to obtain the roots of the desired cubic $f(x)$

We note there are no other cases for the possibilities of the roots of a cubic (i.e. there is no instance where one finds one complex and two real roots). This result is intuitively obvious after graphing cubic polynomials and taking into account that imaginary^{} roots may only occur in conjugate pairs.

Title | trigonometric cubic formula |
---|---|

Canonical name | TrigonometricCubicFormula |

Date of creation | 2013-03-22 15:02:07 |

Last modified on | 2013-03-22 15:02:07 |

Owner | mathcam (2727) |

Last modified by | mathcam (2727) |

Numerical id | 10 |

Author | mathcam (2727) |

Entry type | Theorem |

Classification | msc 12D10 |

Synonym | Alternate cubic formula |

Related topic | CardanosFormulae |