# Proving trigonometric equation $\cos(36^\circ) - \cos(72^\circ) = 1/2$ [duplicate]

$\cos \left( 36^\circ \right)-\cos \left( 72^\circ \right) = \frac{1}{2}$

Thank you.

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## marked as duplicate by Martin Sleziak, Mauro ALLEGRANZA, Omran Kouba, Norbert, 91500Jun 9 '14 at 9:14

This is the same question as: math.stackexchange.com/questions/130817/… – Martin Sleziak Jun 9 '14 at 7:41

As $\displaystyle\cos(180^\circ-y)=-\cos y$

$$S=\cos36^\circ-\cos72^\circ=\cos36^\circ+\cos(180^\circ-108^\circ)=\cos36^\circ+\cos108^\circ$$

Now multiplying the numerator & the denominator by $\displaystyle2\sin\frac{(108-36)}2^\circ,$

$$S=\frac{2\sin36^\circ\cos36^\circ+2\sin36^\circ\cos108^\circ}{2\sin36^\circ}$$

Using Werner Formula & $\sin2x$ formula, $$S=\frac{\sin72^\circ+\sin144^\circ-\sin72^\circ}{2\sin36^\circ}=\frac{\sin(180^\circ-36^\circ)}{2\sin36^\circ}$$

Hope you can take it home from here

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because $sin(180-36)=sin(\pi-\alpha)=sin(\alpha)=sin(36)$ – dato datuashvili Jun 9 '14 at 5:17
Thank you. I appreciate it. – xzatulx Jun 9 '14 at 5:23
@xzatulx, I've added two answers here : math.stackexchange.com/questions/130817/… and math.stackexchange.com/questions/130817/… – lab bhattacharjee Jun 11 '14 at 18:50

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:

$\cos36^{\circ} - \cos72^{\circ} = \dfrac{\sqrt{5}+1}{4} - \dfrac{\sqrt{5}-1}{4} = \dfrac{1}{2}$

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If you could give me the link for the trig functions at the given angles? Thank you. – xzatulx Jun 9 '14 at 5:24
@xzatulx: you can type my username and scroll down the posts and find the one with "trig equations". – DeepSea Jun 9 '14 at 5:27
Or, you can see this: cut-the-knot.org/pythagoras/cos36.shtml – JimmyK4542 Jun 9 '14 at 5:27