Amazing Trigonometry_5

$\cos{6^\circ}\sin{24^\circ}\cos{72^\circ} \\ = \cos{6^\circ} (\sin{6^\circ} \cos{18^\circ} + \cos{6^\circ} \sin{18^\circ}) \cos{72^\circ} \\ = \dfrac{1}{2} [2\cos{6^\circ} \sin{6^\circ} \cos{18^\circ} + (2\cos^2{6^\circ} - 1 +1) \sin{18^\circ} ] \cos{72^\circ} \\ = \dfrac{1}{2} [\sin{12^\circ} \cos{18^\circ} + \cos{12^\circ} \sin{18^\circ} + \sin{18^\circ} ] \cos{72^\circ} \\ = \dfrac{1}{2} [\sin{30^\circ} + \sin{18^\circ} ] \cos{72^\circ} \\ = \dfrac{1}{2} \left(\dfrac{1}{2}+ \sin{18^\circ} \right) \cos{72^\circ} \\= \dfrac{1}{4} \cos{72^\circ} + \dfrac {1}{2} \sin{18^\circ} \sin{(90^\circ-72^\circ)} \\= \dfrac{1}{4} \cos{72^\circ} - \dfrac {1}{4} (1 - 2 \sin^2{18^\circ} - 1) \\= \dfrac{1}{4} \cos{72^\circ} - \dfrac {1}{4} \cos{36^\circ} + \dfrac{1}{4} \\= - \dfrac{1}{4} \cos{108^\circ} - \dfrac {1}{4} \cos{36^\circ} + \dfrac{1}{4} \\= \dfrac{1}{4} [1 - \color{#3D99F6}{(\cos{36^\circ} + \cos{108^\circ})}] \quad \quad \quad \quad \color{#3D99F6}{ \text{[See Note]}}\\ = \dfrac{1}{4} \left[1 - \color{#3D99F6}{\dfrac{1}{2}} \right] = \dfrac {1}{8} \quad \Rightarrow \dfrac {1} {\cos{6^\circ}\sin{24^\circ}\cos{72^\circ}} = \boxed{8}$

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$\color{#3D99F6}{\text{Note: } z^{10} = e^{\frac{\pi}{5}i} = 1 \text{, the 10}^{th} \text{ roots of unity. By Argand's diagram,} \\ \text{we note that } \cos{\frac{\pi}{5}} + \cos{\frac{3\pi}{5}} = \frac{1}{2}.}$

$\begin{array}{c}&\cos6^{\circ}\cos24^{\circ}\cos72^{\circ}&=\dfrac{1}{2}\left(\sin30^{\circ}+\sin18^{\circ}\right)\cos72^{\circ}=\dfrac{1}{4}\cos72^{\circ}+\dfrac{1}{2}\sin18^{\circ}\cos72^{\circ}\\\\&=\dfrac{1}{4}\sin18^{\circ}+\frac{1}{4}\left(\sin90^{\circ}-\sin54^{\circ}\right)=\dfrac{1}{4}+\dfrac{1}{4}\left(\sin18^{\circ}-\sin54^{\circ}\right)\\\\&=\dfrac{1}{4}+\dfrac{1}{4}\dfrac{\left(-2\cos36^{\circ}\sin18^{\circ}\right)\cos18^{\circ}}{\cos18^{\circ}}=\dfrac{1}{4}-\dfrac{1}{4}\dfrac{\left(\cos36^{\circ}\sin36^{\circ}\right)}{\cos18^{\circ}}\\\\&=\dfrac{1}{4}-\dfrac{1}{8}\dfrac{\sin72^{\circ}}{\sin72^{\circ}}=\dfrac{1}{8}\end{array}$

Thus, $\large{\left[\cos6^{\circ}\cos24^{\circ}\cos72^{\circ}\right]^{-1}=8}$

$=\cos6^{\circ}\cos{66}^{\circ}\cos{72}^{\circ}$

$=\dfrac{\cos{72}^{\circ}}{4}\left(4\cos{54}^{\circ}\cos{6}^{\circ}\cos{66}^{\circ}\right)$

$=\dfrac{\cos{72}^{\circ}\cos{18}^{\circ}}{4\cos{54}^{\circ}} \qquad \color{#3D99F6}{\cos(3A) = 4\cos(60-A)\cos(A)\cos(60+A)}$

$=\dfrac{2\sin{18}^{\circ}\cos{18}^{\circ}}{8\sin{36}^{\circ}} \qquad \color{#3D99F6}{\sin(90-A) = \cos(A) \quad , \cos(90-A)=\sin(A)}$

$=\dfrac{\sin{36}^{\circ}}{8\sin{36}^{\circ}} \qquad \qquad \quad \color{#3D99F6}{2\sin(A)\cos(A) = \sin(2A)}$

$=\boxed{\dfrac18}$

$\therefore \text{Our answer is} \quad \boxed{8}$