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.