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Prove that $\dfrac{1}{2}\sqrt{4\sin^2(36^{\circ})-1} = \cos(72^{\circ})$

This question seemed pretty simple so I first started out by turning the left-hand side into terms of $\cos(x)$. We have $\dfrac{1}{2}\sqrt{4(1-\cos^2(36^{\circ}))-1} = \dfrac{1}{2}\sqrt{3-4\cos^2(36^{\circ})}$. Then I would use $\cos(72^{\circ}) = 2\cos^2(36^{\circ})-1$. Thus, $\dfrac{1}{2}\sqrt{3-4\cos^2(36^{\circ})} = \dfrac{1}{2}\sqrt{3-2(\cos(72^{\circ})+1)} = \dfrac{1}{2}\sqrt{1-2\cos(72^{\circ})}$. I get stuck here.

Jacob Willis
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2 Answers2

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Note that we have the system of equations $$\begin{cases}\cos 72^\circ&=2\cos^2 36^\circ-1,\\\cos 36^\circ&=1-2\cos^2 72^\circ.\end{cases}$$ Adding the two equations together yields $$\cos 72^\circ+\cos 36^\circ=2(\cos^236^\circ-\cos^272^\circ)\implies\cos 36^\circ-\cos 72^\circ=\frac12.$$ Now subtracting the first equation from the second gives $$\cos 36^\circ-\cos 72^\circ=2-2(\cos^236^\circ+\cos^272^\circ)\implies\cos^236^\circ+\cos^272^\circ=\frac34.$$ To finish, note that this equality rearranges to $$1-\cos^2 36^\circ-\cos^272^\circ=\sin^236^\circ-\cos^272^\circ=\frac14.$$ Hence $4\sin^236^\circ-1=4\cos^272^\circ$, and taking the square root of both sides and dividing by $2$ gives the desired result.

David Altizio
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  • As a sidenote, the main idea behind these manipulations (computing $\cos 36^\circ-\cos 72^\circ$) is actually AHSME 1975 #30 - remembering the solution to that problem was very helpful in solving this one. – David Altizio Jan 04 '16 at 21:30
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We need $$4\sin^236^\circ-4\cos^272^\circ=1$$

Using $\cos2A=1-2\sin^2A=2\cos^2A-1,$

$$\iff2(1-\cos72^\circ)-2(1+\cos144^\circ)=1$$

As $\cos(180^\circ-B)=-\cos B,$ $$\iff\cos36^\circ-\cos72^\circ=\dfrac12$$

Utilize Proving trigonometric equation $\cos(36^\circ) - \cos(72^\circ) = 1/2$

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lab bhattacharjee
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