Proving trigonometric equation $\cos(36^\circ) - \cos(72^\circ) = 1/2$ Please help me to prove this trigonometric equation.
$\cos \left( 36^\circ \right)-\cos \left( 72^\circ \right) = \frac{1}{2}$
Thank you.

From a recent post which relates to the above trig functions at the given angles, we have:
$\cos36^{\circ} = \sin54^{\circ} = \dfrac{\sqrt{5}+1}{4}$, and:
$\cos72^{\circ} = \sin18^{\circ} = \dfrac{\sqrt{5}-1}{4}$. Thus:
$\cos36^{\circ} - \cos72^{\circ} = \dfrac{\sqrt{5}+1}{4} - \dfrac{\sqrt{5}-1}{4} = \dfrac{1}{2}$
