A trigonometry problem by Veides Kasera

Writing it in $\LaTeX$ , where I'm pretty sure it will benefit somebody:

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We know, $\frac12 + \frac{\sqrt5-1}4 = \frac{\sqrt5+1}4$

$\begin{array} {l c l } (1) & \implies & \sin30^\circ + \sin18^\circ = \sin54^\circ = \cos36^\circ \\ \phantom 0 \\ (2) & \implies & 2\sin 24^\circ \cos 6^\circ = \cos36^\circ \\ \phantom 0 \\ (3) & \implies & \dfrac{\sin24^\circ}{\tfrac12} = \dfrac{\cos36^\circ}{\cos6^\circ} \\ \phantom 0 \\ (4) & \implies & \dfrac{\sin24^\circ}{\sin30^\circ} = \dfrac{\cos36^\circ}{\cos6^\circ} \\ \phantom 0 \\ (5) & \implies & \dfrac{\sin24^\circ + \sin30^\circ}{\sin24^\circ - \sin30^\circ} = \dfrac{\cos36^\circ + \cos6^\circ}{\cos36^\circ - \cos6^\circ} \\ \phantom 0 \\ (6) & \implies & \dfrac{2\sin 27^\circ \cos3^\circ}{-2\sin3^\circ \cos27^\circ} = \dfrac{2\cos21^\circ \cos15^\circ}{-2\sin21^\circ \sin15^\circ} \\ \phantom 0 \\ (7) & \implies & \dfrac{\sin3^\circ}{\cos3^\circ} = \left(\dfrac{\sin27^\circ}{\cos27^\circ} \right) \left(\dfrac{\sin21^\circ}{\cos21^\circ} \right) \left(\dfrac{\sin15^\circ}{\cos15^\circ} \right) \\ \phantom 0 \\ & \therefore & \tan^{-1} \left [\tan21^\circ \tan27^\circ \tan25^\circ \right ] = \boxed{3^\circ} \end{array}$

Values of $\sin18^\circ$ and $\sin54^\circ$ degree are used and can be found in any standard list.

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