Prove that $\cos 20^{\circ} + \cos 100^{\circ} + \cos {140^{\circ}} = 0$ Assume $A = \cos 20^{\circ} + \cos 100^{\circ} + \cos 140^{\circ}$ . Prove that value of $A$ is zero.
My try : $A = 2\cos 60^{\circ} \cos 40^{\circ} + \cos 140^{\circ}$ and I'm stuck here 

$\cos 20^{\circ} + \cos 100^{\circ} = \cos (60^{\circ}-40^{\circ}) + \cos (60^{\circ}+40^{\circ}) = \cos 40^{\circ}$ by using the addition formulae for $\cos$. 
Then $\cos 140^{\circ} = \cos (180^{\circ} - 40^{\circ}) = -\cos (40^{\circ})$. So $$\cos 20^{\circ} + \cos 100^{\circ} + \cos 140^{\circ} = \cos 40^{\circ} - \cos 40^{\circ} = 0$$
