Converging Trig!

Since sin 18 is a constant, we can take it out from summation, also since (cos 18)<1 , $\displaystyle \sum_{n=0}^{\infty} (cos 18)^n$ represents a sum of infinite $GP$ with first term $1$ and common difference $\cos 18$ . $(\sin18) \times \displaystyle \sum_{n=0}^{\infty} (\cos18)^n =\dfrac{\sin 18}{1-(\cos18)}$ $\small({\color{#D61F06}{\text{Using sin2x=2 (sinx)(cosx) and 1-cos2x=2}\sin^2 x})}$ $\large =\dfrac{2(\sin 9)(\cos 9)}{2\sin^2 9}=\cot 9$ Hence, $\Large (9)^{\frac{3}{2}}=\huge\boxed{\color{#007fff}{27}}$

$\begin{aligned} \sum_{n=0}^\infty \sin 18^\circ \cos ^n 18^\circ & = \sin 18^\circ \color{#3D99F6} {\sum_{n=0}^\infty \cos ^n 18^\circ} \quad \quad \small \color{#3D99F6}{\text{As } 0 <\cos 18^\circ < 1 \text{, sum of GP}} \\ & = \sin 18^\circ \color{#3D99F6}{ \left(\frac{1}{1-\cos 18^\circ} \right)} \quad \quad \small \color{#3D99F6}{\text{Using half-angle identities}} \\ & = \frac{\dfrac{2\tan 9^\circ}{1+\tan^2 9^\circ}}{1-\dfrac{1-\tan^2 9^\circ}{1+\tan^2 9^\circ}} \\ & = \frac{2 \tan 9^\circ}{1+\tan^2 9^\circ - 1+\tan^2 9^\circ} \\ & = \frac{2 \tan 9^\circ}{2\tan^2 9^\circ} = \frac{1}{\tan 9^\circ} = \cot 9^\circ \end{aligned}$

$\Rightarrow \beta^{\frac{3}{2}} = 9^{\frac{3}{2}} = \boxed{27}$

$\sum_{n = 0}^{\infty}{ar^n} = \frac{a}{1 - r}$

$a = \sin 18^\circ,~ r = \cos 18^\circ$

$\sum_{n = 0}^{\infty}{\sin 18^\circ} (\cos 18^\circ)^n = \frac{\sin 18^\circ}{1 - \cos 18^\circ}$

$\frac{\sin 18^\circ}{1 - \cos 18^\circ} = \cot \beta$

$\frac{\sin 18^\circ}{1 - \cos 18^\circ} = \frac{1}{\tan \beta}$

$\frac{1 - \cos 18^\circ}{\sin 18^\circ} = \tan \beta$

$\frac{1 - (\cos^2 9^\circ - \sin^2 9^\circ)}{\sin 18^\circ} = \tan \beta$

$\frac{\sin^2 9^\circ + \cos^2 9^\circ + \sin^2 9^\circ - \cos^2 9^\circ}{\sin 18^\circ} = \tan \beta$

$\frac{2\sin^2 9^\circ}{\sin 18^\circ} = \tan \beta$

$\frac{2\sin^2 9^\circ}{2\sin 9^\circ\cos 9^\circ} = \tan \beta$

$\frac{\sin 9^\circ}{cos 9^\circ} = \tan \beta$

$\tan 9^\circ = \tan \beta$

$\beta = 9$

$\beta^\frac{3}{2} = 9^\frac{3}{2} = 27$

$\large \boxed{\beta^\frac{3}{2} = 27}$