A geometry problem by Aakhyat Singh

$\begin{aligned} X & = \cos 40^\circ + \cos 80^\circ + {\color{#3D99F6}\cos 160^\circ} + {\color{#3D99F6}\cos 240^\circ} \quad \quad \quad \quad \small \color{#3D99F6} \text{Note that } \cos (180^\circ - x) = - \cos x \\ & = \cos (60^\circ - 20^\circ) + \cos (60^\circ + 20^\circ) \color{#3D99F6} - \cos 20^\circ - \cos 60^\circ \\ & = \cos 60^\circ \cos 20^\circ + \sin 60^\circ \sin 20^\circ + \cos 60^\circ \cos 20^\circ - \sin 60^\circ \sin 20^\circ - \cos 20^\circ - \cos 60^\circ \\ & = 2 \cos 60^\circ \cos 20^\circ - \cos 20^\circ - \cos 60^\circ \\ & = 2 \times \frac 12 \times \cos 20^\circ - \cos 20^\circ - \frac 12 \\ & = - \frac 12 = \boxed{-0.5} \end{aligned}$

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Alternative Solution

$\begin{aligned} X & = \cos 40^\circ + \cos 80^\circ + \cos 160^\circ + \cos 240^\circ \\ & = \cos \frac {2\pi}9 + \cos \frac {4\pi}9 + \cos \frac {8\pi}9 + \color{#3D99F6}\cos \frac {12\pi}9 & \small \color{#3D99F6} \text{Note that }\cos (2\pi -x) = \cos x \\ & = \cos \frac {2\pi}9 + \cos \frac {4\pi}9 + {\color{#3D99F6}\cos \frac {6\pi}9} + \cos \frac {8\pi}9 & \small \color{#3D99F6} \text{See note: }\sum_{k=0}^{n-1} \cos \left(\frac {2k}{2n+1}\pi \right) = - \frac 12 \\ & = - \frac 12 = \boxed{-0.5} \end{aligned}$

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Note: See here for the proof for $\displaystyle \sum_{k=0}^{n-1} \cos \left(\frac {2k}{2n+1}\pi \right) = - \frac 12$ .