Cardano's Formula/Trigonometric Form

Theorem

Let $P$ be the cubic equation:

$a x^3 + b x^2 + c x + d = 0$ with $a \ne 0$

Let:

$Q = \dfrac {3 a c - b^2} {9 a^2}$
$R = \dfrac {9 a b c - 27 a^2 d - 2 b^3} {54 a^3}$

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:

$x_1 = 2 \sqrt {-Q} \map \cos {\dfrac \theta 3} - \dfrac b {3 a}$
$x_2 = 2 \sqrt {-Q} \map \cos {\dfrac \theta 3 + \dfrac {2 \pi} 3} - \dfrac b {3 a}$
$x_3 = 2 \sqrt {-Q} \map \cos {\dfrac \theta 3 + \dfrac {4 \pi} 3} - \dfrac b {3 a}$

where:

$\cos \theta = \dfrac R {\sqrt{-Q^3} }$

Proof

From Cardano's Formula, the roots of $P$ are:

$(1): \quad x_1 = S + T - \dfrac b {3 a}$
$(2): \quad x_2 = - \dfrac {S + T} 2 - \dfrac b {3 a} + \dfrac {i \sqrt 3} 2 \paren {S - T}$
$(3): \quad x_3 = - \dfrac {S + T} 2 - \dfrac b {3 a} - \dfrac {i \sqrt 3} 2 \paren {S - T}$

where:

$S = \sqrt [3] {R + \sqrt {Q^3 + R^2} }$
$T = \sqrt [3] {R - \sqrt {Q^3 + R^2} }$

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:

$(1): \quad x_1 = 2 \sqrt {-Q} \, \map \cos {\dfrac \theta 3} - \dfrac b {3 a}$
$(2): \quad x_2 = 2 \sqrt {-Q} \, \map \cos {\dfrac \theta 3 + \dfrac {2 \pi} 3} - \dfrac b {3 a}$
$(3): \quad x_3 = 2 \sqrt {-Q} \, \map \cos {\dfrac \theta 3 + \dfrac {4 \pi} 3} - \dfrac b {3 a}$

follows after some algebra.

$\blacksquare$

Historical Note

The trigonometric form of Cardano's Formula was devised by François Viète and published in $1591$.