MHB Harrison's question via Facebook about polar functions

[Prove It]

The point $\displaystyle \begin{align*} \left( \rho , \alpha \right) \end{align*}$ lies on the curve $\displaystyle \begin{align*} r = \frac{3\,\theta}{2} \end{align*}$ and the point $\displaystyle \begin{align*} \left( \rho , \beta \right) \end{align*}$ lies on the curve $\displaystyle \begin{align*} r = \theta + \pi \end{align*}$, such that the points are the same distance from the origin, $\displaystyle \begin{align*} 0\leq \theta \leq \pi \end{align*}$ and the distance between them is $\displaystyle \begin{align*} \sqrt{3}\,\pi \end{align*}$. Show that $\displaystyle \begin{align*} \alpha \end{align*}$ satisfies $\displaystyle \begin{align*} \frac{2\,\pi^2}{3} = \alpha ^2 \,\left[ 1 + \cos{ \left( \frac{\alpha}{2} \right) } \right] \end{align*}$

Since the distances from the origin $\displaystyle \begin{align*} \rho \end{align*}$ are the same, we can say $\displaystyle \begin{align*} \rho = \frac{3\,\alpha}{2} \end{align*}$ and $\displaystyle \begin{align*} \rho = \beta + \pi \end{align*}$, giving

$\displaystyle \begin{align*} \frac{3\,\alpha}{2} &= \beta + \pi \\ \beta &= \frac{3\,\alpha}{2} - \pi \end{align*}$

The distance between two points in polar form $\displaystyle \begin{align*} \left( r_1 , \theta_1 \right) \end{align*}$ and $\displaystyle \begin{align*} \left( r_2, \theta_2 \right) \end{align*}$ is given by $\displaystyle \begin{align*} d = \sqrt{r_1^2 + r_2^2 - 2\,r_1\,r_2\,\cos{ \left( \theta_1 - \theta_2 \right) }} \end{align*}$, so in this case

$\displaystyle \begin{align*} \sqrt{3}\,\pi &= \sqrt{ \rho^2 + \rho^2 - 2\,\rho^2 \,\cos{ \left( \alpha - \beta \right) } } \\ \sqrt{3}\,\pi &= \sqrt{ 2\,\rho^2 \,\left[ 1 - \cos{ \left( \alpha - \beta \right) } \right] } \\ 3\,\pi^2 &= 2\,\rho ^2 \,\left[ 1 - \cos{ \left( \alpha - \beta \right) } \right] \\ 3\,\pi^2 &= 2\,\left( \frac{3\,\alpha}{2} \right) ^2 \,\left\{ 1 - \cos{ \left[ \alpha - \left( \frac{3\,\alpha}{2} - \pi \right) \right] } \right\} \\ 3\,\pi^2 &= \frac{9\,\alpha^2}{2} \,\left[ 1 - \cos{\left( \pi - \frac{\alpha}{2} \right) } \right] \\ \frac{2\,\pi^2}{3} &= \alpha^2 \,\left[ 1 - \cos{\left( \pi - \frac{\alpha}{2} \right) } \right] \\ \frac{2\,\pi^2}{3} &= \alpha^2\,\left[ 1 + \cos{ \left( \frac{\alpha}{2} \right) } \right] \end{align*}$

 

[Mark44]

QED