Trigonometric formula for solving quadratic equations

In this note, we provide an alternative trigonometric formula for solving quadratic equations where $a$, $b$, and $c$ are non-zero real numbers.

If $ax^2+bx+c=0$, then 

$$x_{1,2}=\left(1\pm\frac{2}{\tan{\frac{\theta}{2}\mp1}}\right)\sqrt{\frac{c}{a}},$$

where $\theta=\sec^{-1}{\left(\frac{b}{2\sqrt{ac}}\right)}$.

Proof. Multiplying by $a$ the quadratic equation we have

$$(ax)^2+b(ax)+ac=0.$$

Let's make the substitution $\sec{\theta}=\frac{b}{2\sqrt{ac}}$, then

$$\begin{aligned}0&=(ax)^2+b(ax)+ac\\&=(ax)^2+2\sqrt{ac}(ax)\sec{\theta}+ac\\&=\sec{\theta}\left((ax)^2\cos{\theta}+2\sqrt{ac}(ax)+ac\cos{\theta}\right)\\&=\cos{\theta}((ax)^2+ac)+2\sqrt{ac}(ax)\\&= \cos{\theta}(ax+\sqrt{ac})^2-2\sqrt{ac}(ax)\cos{\theta}+2\sqrt{ac}(ax)\\&= \left(\cos^2{\frac{\theta}{2}}-\sin^2{\frac{\theta}{2}}\right)(ax+\sqrt{ac})^2-2\sqrt{ac}(ax)\left(1-2\sin^2{\frac{\theta}{2}}\right)+2\sqrt{ac}(ax)\\&=  \cos^2{\frac{\theta}{2}}(ax+\sqrt{ac})^2 -\sin^2{\frac{\theta}{2}}(ax+\sqrt{ac})^2 +4\sqrt{ac}(ax)\sin^2{\frac{\theta}{2}}\\&=   \cos^2{\frac{\theta}{2}}(ax+\sqrt{ac})^2 -\sin^2{\frac{\theta}{2}}\left((ax+\sqrt{ac})^2 -4\sqrt{ac}(ax)\right)\\&= \cos^2{\frac{\theta}{2}}(ax+\sqrt{ac})^2 -\sin^2{\frac{\theta}{2}}(ax-\sqrt{ac})^2\\&= \left(\cos{\frac{\theta}{2}}(ax+\sqrt{ac}) +\sin{\frac{\theta}{2}}(ax-\sqrt{ac})\right) \left(\cos{\frac{\theta}{2}}(ax+\sqrt{ac}) -\sin{\frac{\theta}{2}}(ax-\sqrt{ac})\right)\\&=\left(ax\left(\cos{\frac{\theta}{2}}+\sin{\frac{\theta}{2}}\right)+\sqrt{ac}\left(\cos{\frac{\theta}{2}}-\sin{\frac{\theta}{2}}\right)  \right) \left(ax\left(\cos{\frac{\theta}{2}}-\sin{\frac{\theta}{2}}\right)+\sqrt{ac}\left(\cos{\frac{\theta}{2}}+\sin{\frac{\theta}{2}}\right)  \right).\end{aligned}$$

Now, by setting the factors equal to zero and solving for x, we obtain,

$$\begin{aligned}x_1&=\left(\frac{\sin{\frac{\theta}{2}}+\cos{\frac{\theta}{2}}}{\sin{\frac{\theta}{2}}-\cos{\frac{\theta}{2}}}\right)\sqrt{\frac{c}{a}}\\&= \left(1+\frac{2}{\tan{\frac{\theta}{2}-1}}\right)\sqrt{\frac{c}{a}}. \end{aligned}$$

Similarly we obtain, 

$$\begin{aligned}x_2= \left(1-\frac{2}{\tan{\frac{\theta}{2}+1}}\right)\sqrt{\frac{c}{a}} \end{aligned}.$$

$\square$

The formula appears to be new, at least on the internet. Wikipedia presents a trigonometric method, but it is different from mine. Stuart Simons offers a method that does seem to be related to my approach, but he only provides direct formulas for complex roots. I was able to independently derive Simons' formulas using $\cos{\theta}$ in the substitution instead of $\sec{\theta}$, and this was before coming across the Wikipedia article that references Simons' paper: Simons, Stuart, 'Alternative approach to complex roots of real quadratic equations,' Mathematical Gazette 93, March 2009, 91–92.

More discussion about this formula can be found on the MathSE forum.

Remark. The half-angle formulas are ubiquitous, and this alternative formula for quadratic equations is another piece of evidence for that.

A generalization of Burlet's theorem to cyclic quadrilaterals

 The Burlet's theorem is a result in Euclidean geometry, which can be formulated as follows:

Theorem 1. Consider triangle $ABC$ with $\angle{BCA}=\gamma$. Let $P$ be the point where the incircle touches side $AB$, and denote the lengths $AP$ and $BP$ as $m$ and $n$, respectively. Then

$$\Delta_0=mn\cot{\frac{\gamma}{2}},\tag{1}$$

A triangle $ABC$ with $AP=m$ and $BP=n$.

where $\Delta_0$ denotes the area of $ABC$. Note that when $\gamma=\frac{\pi}{2}$, then $\Delta_0=mn$.

We adopt unconventional notation to reduce the relation of Theorem 2 into the one referenced in $(1)$. Additionally, we denote $s$ the semiperimeter, applicable to both triangles and cyclic quadrilaterals.

Proof. Let $AB=a$, $BC=b$ and $AC=c$. By Heron's formula, we have

$$\begin{aligned}\Delta_0&=\sqrt{s(s-a)(s-b)(s-c)}\\&=\sqrt{s(s-a)(s-b)(s-c)}\color{red}{\sqrt{\frac{(s-b)(s-c)}{(s-b)(s-c)}}}\\&=(s-b)(s-c)\sqrt{\frac{s(s-a)}{(s-b)(s-c)}}.\end{aligned}$$

However, using half-angle formulas for a triangle, we can express $\cot{\frac{\gamma}{2}}$ as $\sqrt{\frac{s(s-a)}{(s-b)(s-c)}}$, where $m=(s-b)$ and $n=(s-c)$. This establishes the relationship mentioned in $(1)$.

$\square$

Theorem 2 (Generalization). Consider a cyclic quadrilateral $ABCD$ with side lengths $AB=a$, $BC=b$, $CD=c$, and $DA=d$. Additionally, let $\angle DAB = \alpha$. In this context, the following relation is valid:

$$\Delta_1=(s-a)(s-d)\cot{\frac{\alpha}{2}},\tag{2}$$

A cyclic quadrilateral $ABCD$.

where $\Delta_1$ denotes the area of $ABCD$.

Proof. Similar to how we proceeded in theorem 1, using the Brahmagupta's formula, we have

$$\begin{aligned}\Delta_1&=\sqrt{(s-a)(s-b)(s-c)(s-d)}\\&=\sqrt{(s-a)(s-b)(s-c)(s-d)}\color{red}{\sqrt{\frac{(s-a)(s-d)}{(s-a)(s-d)}}}\\&=(s-a)(s-d)\sqrt{\frac{(s-b)(s-c)}{(s-a)(s-d)}}.\end{aligned}$$

Now, invoking the half-angle formulas for cyclic quadrilaterals, we have that $\cot{\frac{\alpha}{2}}=\sqrt{\frac{(s-b)(s-c)}{(s-a)(s-d)}}$, from which the relation $(2)$ is deduced.

$\square$

Let $\angle{BCD}=\gamma$ and assume that $d=0$. We have that $\alpha=\pi-\gamma$, so substituting in $(2)$, 

When $d=0$, by the alternate segment theorem, $\alpha=\pi-\gamma$.

$$\begin{aligned}\Delta_1&=(s-a)(s-0)\cot{\left(\frac{\pi-\gamma}{2}\right)}\\&=s(s-a)\tan{\frac{\gamma}{2}}\\&=s(s-a)\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}\color{red}{\sqrt{\frac{(s-b)(s-c)}{(s-b)(s-c)}}}\\&=(s-b)(s-c)\sqrt{\frac{s(s-a)}{(s-b)(s-c)}}\\&=(s-b)(s-c)\cot{\frac{\gamma}{2}},\end{aligned}$$

which is Burlet's theorem for a triangle since $m=(s-b)$ and $n=(s-c)$.

Remark. A. Burlet (Dublin) presented this theorem as a problem in the Nouvelles annales de mathématiques, Vol. 15, 1856, p. 290. Consequently, the theorem might have derived its name from him.