Cardano's Formula/Trigonometric Form
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Theorem
Let $P$ be the cubic equation:
- $a x^3 + b x^2 + c x + d = 0$ with $a \ne 0$
Let:
\(\ds Q\) | \(=\) | \(\ds \dfrac {3 a c - b^2} {9 a^2}\) | ||||||||||||
\(\ds R\) | \(=\) | \(\ds \dfrac {9 a b c - 27 a^2 d - 2 b^3} {54 a^3}\) |
Let $a, b, c, d \in \R$.
Let the discriminant $D < 0$, where $D := Q^3 + R^2$.
Then the solutions of $P$ can be expressed as:
\(\text {(1)}: \quad\) | \(\ds x_1\) | \(=\) | \(\ds 2 \sqrt {-Q} \map \cos {\dfrac \theta 3} - \dfrac b {3 a}\) | |||||||||||
\(\text {(2)}: \quad\) | \(\ds x_2\) | \(=\) | \(\ds 2 \sqrt {-Q} \map \cos {\dfrac \theta 3 + \dfrac {2 \pi} 3} - \dfrac b {3 a}\) | |||||||||||
\(\text {(3)}: \quad\) | \(\ds x_3\) | \(=\) | \(\ds 2 \sqrt {-Q} \map \cos {\dfrac \theta 3 + \dfrac {4 \pi} 3} - \dfrac b {3 a}\) |
where:
- $\cos \theta = \dfrac R {\sqrt {-Q^3} }$
Proof
From Cardano's Formula, the roots of $P$ are:
\(\text {(1)}: \quad\) | \(\ds x_1\) | \(=\) | \(\ds S + T - \dfrac b {3 a}\) | |||||||||||
\(\text {(2)}: \quad\) | \(\ds x_2\) | \(=\) | \(\ds -\dfrac {S + T} 2 - \dfrac b {3 a} + \dfrac {i \sqrt 3} 2 \paren {S - T}\) | |||||||||||
\(\text {(3)}: \quad\) | \(\ds x_3\) | \(=\) | \(\ds -\dfrac {S + T} 2 - \dfrac b {3 a} - \dfrac {i \sqrt 3} 2 \paren {S - T}\) |
where:
\(\ds S\) | \(=\) | \(\ds \sqrt [3] {R + \sqrt {Q^3 + R^2} }\) | ||||||||||||
\(\ds T\) | \(=\) | \(\ds \sqrt [3] {R - \sqrt {Q^3 + R^2} }\) |
Let $D = Q^3 + R^2 < 0$.
Then:
- $S^3 = R + i \sqrt {\size {Q^3 + R^2} }$
We can express this in polar form:
- $S^3 = r \paren {\cos \theta + i \sin \theta}$
where:
- $r = \sqrt {R^2 + \paren {\sqrt {Q^3 + R^2} }^2} = \sqrt {R^2 - \paren {Q^3 + R^2} } = \sqrt {-Q^3}$
- $\tan \theta = \dfrac {\sqrt {\size {Q^3 + R^2} } } R$
Then:
- $\cos \theta = \dfrac R {\sqrt {-Q^3} }$
Similarly for $T^3$.
The result:
\(\text {(1)}: \quad\) | \(\ds x_1\) | \(=\) | \(\ds 2 \sqrt {-Q} \map \cos {\dfrac \theta 3} - \dfrac b {3 a}\) | |||||||||||
\(\text {(2)}: \quad\) | \(\ds x_2\) | \(=\) | \(\ds 2 \sqrt {-Q} \map \cos {\dfrac \theta 3 + \dfrac {2 \pi} 3} - \dfrac b {3 a}\) | |||||||||||
\(\text {(3)}: \quad\) | \(\ds x_3\) | \(=\) | \(\ds 2 \sqrt {-Q} \map \cos {\dfrac \theta 3 + \dfrac {4 \pi} 3} - \dfrac b {3 a}\) |
follows after some algebra.
$\blacksquare$
Historical Note
The Trigonometric Form of Cardano's Formula was devised by François Viète and published in $1591$.
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 9$: Solutions of Algebraic Equations: Cubic Equation: $x^3 + a_1 x^2 + a_2 x + a_3 = 0$: $9.4$
- 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): cubic
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): cubic
- 2009: Murray R. Spiegel, Seymour Lipschutz and John Liu: Mathematical Handbook of Formulas and Tables (3rd ed.) ... (previous) ... (next): $\S 5$: Solutions of Algebraic Equations: Cubic Equation: $x^3 + a_1 x^2 + a_2 x + a_3 = 0$: $5.4.$