This note presents alternative trigonometric formulas for finding the roots of quadratic equations where $a$, $b$, and $c$ are non-zero real numbers.
Before the era of calculators, trigonometric formulas were favored for computing quadratic roots due to their time and labor-saving benefits. However, it's not advisable nowadays to rely on these formulas over the traditional quadratic formula for root calculations. Nonetheless, the trigonometric formulas showcased in this note have hinted at certain identities that appear useful for integrating trigonometric functions. Thus, the sole purpose of this note is to document the origins of these identities.
Theorem 1. Let $a$, $b$, and $c$ be non-zero real numbers. For the quadratic equation $ax^2+bx+c=0$, the roots are given by:
$$x_{1}= ie^{i\alpha}\sqrt{\frac{c}{a}}\qquad \text{and}\qquad x_2=-ie^{-i\alpha}\sqrt{\frac{c}{a}},\tag{1}$$
where $\alpha=\sin^{-1}{\left(\frac{b}{2\sqrt{ac}}\right)}$.
Proof. Consider the quadratic equation $a^2+bx+c=0$. Multiplying both sides by $a$ and making the substitutions $ax=p$ and $ac=q^2$ yields
$$p^2+bp+q^2=0.$$
Let's make the substitution $\sin{\alpha} = \frac{b}{2q}$. Consequently,
$$\begin{aligned}0&=p^2+bp+q^2\\&=p^2+2qp\sin{\alpha}+q^2\\&=p^2\left(\sin^2{\frac{\alpha}{2}}+\cos^2{\frac{\alpha}{2}}\right)+4qp\sin{\frac{\alpha}{2}}\cos{\frac{\alpha}{2}}+q^2\left(\sin^2{\frac{\alpha}{2}}+\cos^2{\frac{\alpha}{2}}\right)\\&=\left(p\sin{\frac{\alpha}{2}}+q\cos{\frac{\alpha}{2}}\right)^2 + \left(p\cos{\frac{\alpha}{2}}+q\sin{\frac{\alpha}{2}}\right)^2\\&=\left(p\sin{\frac{\alpha}{2}}+q\cos{\frac{\alpha}{2}}\right)^2 -i^2\left(p\cos{\frac{\alpha}{2}}+q\sin{\frac{\alpha}{2}}\right)^2 .\end{aligned}$$
Factorizing the difference of squares and then factorizing again, one of the factor of the quadratic equation can be express as follows
$$p\left(\sin{\frac{\alpha}{2}}+i\cos{\frac{\alpha}{2}}\right)+q\left(\cos{\frac{\alpha}{2}}+i\sin{\frac{\alpha}{2}}\right).$$
Setting the factor equal to zero, undoing the substitutions $ax=p$ and $ac=q^2$ and solving for $x$, we obtain
$$\begin{aligned}x_1&=-\left(\frac{\cos{\frac{\alpha}{2}}+i\sin{\frac{\alpha}{2}}}{\sin{\frac{\alpha}{2}}+i\cos{\frac{\alpha}{2}}}\right)\sqrt{\frac{c}{a}}\\&= -(\sin{\alpha}-i\cos{\alpha})\sqrt{\frac{c}{a}}\\&=ie^{i\alpha}\sqrt{\frac{c}{a}}. \end{aligned}$$
The other factor is given by
$$p\left(\sin{\frac{\alpha}{2}}-i\cos{\frac{\alpha}{2}}\right)+q\left(\cos{\frac{\alpha}{2}}-i\sin{\frac{\alpha}{2}}\right).$$
And similarly,
$$x_2=-ie^{-i\alpha}\sqrt{\frac{c}{a}}.$$
Theorem 2. For the quadratic equation $ax^2+bx+c=0$ with non-zero real numbers $a$, $b$, and $c$, the roots are given by:
$$x_{1}=-\tan{\frac{\beta}{2}}\sqrt{\frac{c}{a}}\qquad \text{and} \qquad x_{2}=-\cot{\frac{\beta}{2}}\sqrt{\frac{c}{a}},\tag{2}$$
where $\beta=\csc^{-1}{\left(\frac{b}{2\sqrt{ac}}\right)}$.
Proof. Multiplying by $a$ the quadratic equation $ax^2+bx+c=0$ and then substituting $ax=p$ and $ac=q^2$, we have
$$p^2+bp+q^2=0.$$
Use the substitution $\csc{\beta}=\frac{b}{2q}$, then
$$\begin{aligned}0&=p^2+bp+q^2\\&=p^2+2qp\csc{\beta}+q^2\\&=\csc{\beta}\left(p^2\sin{\beta}+2qp+q^2\sin{\beta}\right)\\&=2p^2\sin{\frac{\beta}{2}}\cos{\frac{\beta}{2}}+2qp\left(\sin^2{\frac{\beta}{2}}+\cos^2{\frac{\beta}{2}}\right)+2q^2\sin{\frac{\beta}{2}}\cos{\frac{\beta}{2}}\\&=p^2\sin{\frac{\beta}{2}}\cos{\frac{\beta}{2}}+qp\sin^2{\frac{\beta}{2}}+ qp\cos^2{\frac{\beta}{2}}+q^2\sin{\frac{\beta}{2}}\cos{\frac{\beta}{2}}\\&= p\sin{\frac{\beta}{2}}\left(p\cos{\frac{\beta}{2}}+q\sin{\frac{\beta}{2}}\right)+q\cos{\frac{\beta}{2}}\left(p\cos{\frac{\beta}{2}}+q\sin{\frac{\beta}{2}}\right)\\&= \left(p\cos{\frac{\beta}{2}}+q\sin{\frac{\beta}{2}}\right)\left(p\sin{\frac{\beta}{2}}+q\cos{\frac{\beta}{2}}\right).\end{aligned}$$By setting the factors equal to zero, undoing substitutions $ax=p$ and $ac=q^2$ and solving for $x$, we obtain the desired formulas in $(2)$.
Theorem 3. Let $a$, $b$ and $c$ be non-zero real numbers. If $ax^2+bx+c=0$, then the roots are given by:
$$x_{1}= -e^{i\gamma}\sqrt{\frac{c}{a}}\qquad \text{and}\qquad x_2=-e^{-i\gamma}\sqrt{\frac{c}{a}},\tag{3}$$
where $\gamma=\cos^{-1}{\left(\frac{b}{2\sqrt{ac}}\right)}$.
Proof. Consider the quadratic equation $ax^2+bx+c=0$. Multiplying both sides by $a$ and substituting $ax=p$ and $ac=q^2$ yields
$$p^2+bp+q^2=0.$$
By making the substitution $\cos{\gamma} = \frac{b}{2q}$, it follows that
$$\begin{aligned}0&=p^2+bp+q^2\\&=p^2+2qp\cos{\gamma}+q^2\\&=p^2\left(\sin^2{\frac{\gamma}{2}}+\cos^2{\frac{\gamma}{2}}\right)+4qp\left(\cos^2{\frac{\gamma}{2}}-\sin^2{\frac{\gamma}{2}}\right)+q^2\left(\sin^2{\frac{\gamma}{2}}+\cos^2{\frac{\gamma}{2}}\right)\\&=\sin^2{\frac{\gamma}{2}} (p-q)^2+\cos^2{\frac{\gamma}{2}}(p+q)^2\\&=\sin^2{\frac{\gamma}{2}} (p-q)^2-i^2\cos^2{\frac{\gamma}{2}}(p+q)^2.\end{aligned}$$
By factoring the difference of squares first and then applying further factorization, one of the factors of the quadratic equation can be represented as follows
$$p\left(\sin{\frac{\gamma}{2}}+i\cos{\frac{\gamma}{2}}\right)+q\left(-\sin{\frac{\gamma}{2}}+i\cos{\frac{\gamma}{2}}\right).$$
Setting the factors equal to zero, undoing substitutions $ax=p$ and $ac=q^2$ and solving for $x$, we obtain
$$\begin{aligned}x_1&=\left(\frac{\sin{\frac{\gamma}{2}}-i\cos{\frac{\gamma}{2}}}{\sin{\frac{\gamma}{2}}+i\cos{\frac{\gamma}{2}}}\right)\sqrt{\frac{c}{a}}\\&= -(\cos{\gamma}+i\sin{\gamma})\sqrt{\frac{c}{a}}\\&=-e^{i\gamma}\sqrt{\frac{c}{a}}.\end{aligned}$$
The other factor is given by
$$p\left(\sin{\frac{\gamma}{2}}-i\cos{\frac{\gamma}{2}}\right)-q\left(\sin{\frac{\gamma}{2}}+i\cos{\frac{\gamma}{2}}\right).$$
And similarly,
$$x_2=-e^{-i\gamma}\sqrt{\frac{c}{a}}.$$
Theorem 4. Let $a$, $b$ and $c$ be non-zero real numbers. If $ax^2+bx+c=0$, then the roots are given by:
$$x_{1,2}=\frac{\tan{\frac{\delta}{2}\pm1}}{\tan{\frac{\delta}{2}\mp1}}\sqrt{\frac{c}{a}},\tag{4}$$
where $\delta=\sec^{-1}{\left(\frac{b}{2\sqrt{ac}}\right)}$.
Proof. Consider the quadratic equation $ax^2+bx+c=0$. Multiplying both sides by $a$ and substituting $ax=p$ and $ac=q^2$ yields
$$p^2+bp+q^2=0.$$
Use the substitution $\sec{\delta}=\frac{b}{2q}$, then
$$\begin{aligned}0&=p^2+bp+q^2\\&=p^2+2qp\sec{\delta}+q^2\\&=\sec{\delta}\left(p^2\cos{\delta}+2qp+q^2\cos{\delta}\right)\\&=\cos{\delta}(p^2+q^2)+2qp\\&= \cos{\delta}(p+q)^2-2qp\cos{\delta}+2qp\\&= \left(\cos^2{\frac{\delta}{2}}-\sin^2{\frac{\delta}{2}}\right)(p+q)^2-2qp\left(1-2\sin^2{\frac{\delta}{2}}\right)+2qp\\&= \cos^2{\frac{\delta}{2}}(p+q)^2 -\sin^2{\frac{\delta}{2}}(p+q)^2 +4qp\sin^2{\frac{\delta}{2}}\\&= \cos^2{\frac{\delta}{2}}(p+q)^2 -\sin^2{\frac{\delta}{2}}\left((p+q)^2 -4qp\right)\\&= \cos^2{\frac{\delta}{2}}(p+q)^2 -\sin^2{\frac{\delta}{2}}(p-q)^2\\&= \left(\cos{\frac{\delta}{2}}(p+q) +\sin{\frac{\delta}{2}}(p-q)\right) \left(\cos{\frac{\delta}{2}}(p+q) -\sin{\frac{\delta}{2}}(p-q)\right).\end{aligned}$$
Expanding and then factorizing again, one of the factors of the quadratic equation is given by
$$p\left(\cos{\frac{\delta}{2}}+\sin{\frac{\delta}{2}}\right)+q\left(\cos{\frac{\delta}{2}}-\sin{\frac{\delta}{2}}\right).$$
By setting the factors equal to zero, undoing substitutions $ax=p$ and $ac=q^2$ and solving for $x$, we obtain
$$\begin{aligned}x_1&=\frac{\sin{\frac{\delta}{2}}+\cos{\frac{\delta}{2}}}{\sin{\frac{\delta}{2}}-\cos{\frac{\delta}{2}}}\sqrt{\frac{c}{a}}\\&= \frac{\tan{\frac{\delta}{2}+1}}{\tan{\frac{\delta}{2}-1}}\sqrt{\frac{c}{a}}. \end{aligned}$$
The other factor is given by
$$p\left(\cos{\frac{\delta}{2}}-\sin{\frac{\delta}{2}}\right)+q\left(\cos{\frac{\delta}{2}}+\sin{\frac{\delta}{2}}\right).$$
Similarly,
$$x_2=\frac{\tan{\frac{\delta}{2}-1}}{\tan{\frac{\delta}{2}+1}}\sqrt{\frac{c}{a}}.$$
Theorem 5. Let $a$, $b$ and $c$ be non-zero real numbers. If $ax^2+bx+c=0$, then the roots are given by:
$$x_{1,2}=\frac{i\pm e^{\eta}}{i\mp e^{\eta}}\sqrt{\frac{c}{a}},\tag{5}$$
where $\eta=\tanh^{-1}{\left(\frac{b}{2\sqrt{ac}}\right)}$.
Proof. Consider the quadratic equation $ax^2+bx+c=0$. Multiplying both sides by $a$ and substituting $ax=p$ and $ac=q^2$ yields
$$p^2+bp+q^2=0.$$
Use the substitution $\tanh{\eta}=\frac{b}{2q}$, then
$$\begin{aligned}0&=p^2+bp+q^2\\&=p^2+2qp\tanh{\eta}+q^2\\&=(e^{2\eta} + 1) p^2 + 2 (e^{2\eta} - 1) p q + (e^{2\eta} + 1) q^2\\&=e^{2\eta}(p + q)^2 + (p - q)^2\\&=e^{2\eta}(p + q)^2 - i^2(p - q)^2 \\&=(e^{\eta}p + i p + e^{\eta}q - i q)(e^{\eta}p - i p + e^{\eta}q + i q)\\&= \left(p(e^{\eta} + i) + q(e^{\eta} - i)\right)\left(p(e^{\eta} - i) + q(e^{\eta} + i)\right)\end{aligned}$$
Setting the factor equal to zero, undoing the substitutions $ax=p$ and $ac=q^2$ and solving for $x$, we obtain
$$x_{1}=\frac{i- e^{\eta}}{i+e^{\eta}}\sqrt{\frac{c}{a}}.$$
And similarly,
$$x_{2}=\frac{i+ e^{\eta}}{i-e^{\eta}}\sqrt{\frac{c}{a}}.$$
Theorem 6. Let $a$, $b$ and $c$ be non-zero real numbers. If $ax^2+bx+c=0$, then the roots are given by:
$$x_{1,2}=\frac{1\pm e^{\theta}}{1\mp e^{\theta}}\sqrt{\frac{c}{a}},\tag{6}$$
where $\theta=\coth^{-1}{\left(\frac{b}{2\sqrt{ac}}\right)}$.
Proof. Consider the quadratic equation $ax^2+bx+c=0$. Multiplying both sides by $a$ and substituting $ax=p$ and $ac=q^2$ yields
$$p^2+bp+q^2=0.$$
Use the substitution $\coth{\theta}=\frac{b}{2q}$, Then, similar to how we proceeded previously,
$$\begin{aligned}0&=p^2+bp+q^2\\&=p^2+2qp\coth{\theta}+q^2\\&=(e^{2\theta} - 1) p^2 + 2 (e^{2\theta} + 1) p q + (e^{2\theta} - 1) q^2\\&=e^{2\theta}(p + q)^2 - (p - q)^2 \\&=\left( p e^{\theta} + p + q e^{\theta} - q \right) \left( p e^{\theta} - p + q e^{\theta} + q \right)\\&=\left(p(e^x + 1) + q(e^x - 1)\right)\left(p (e^x - 1) + q (e^x + 1)\right)\end{aligned}$$
Setting the factor equal to zero, undoing the substitutions $ax=p$ and $ac=q^2$ and solving for $x$, we obtain
$$x_{1}=\frac{1- e^{\theta}}{1+e^{\theta}}\sqrt{\frac{c}{a}}= -\tanh\left(\frac{1}{2} \text{coth}^{-1}(x)\right)\sqrt{\frac{c}{a}}.$$
And similarly,
$$x_{2}=\frac{1+ e^{\theta}}{1-e^{\theta}}\sqrt{\frac{c}{a}}.$$