1975 AHSME Problems/Problem 30

Problem 30

Let $x=\cos 36^{\circ} - \cos 72^{\circ}$. Then $x$ equals

$\textbf{(A)}\ \frac{1}{3}\qquad \textbf{(B)}\ \frac{1}{2} \qquad \textbf{(C)}\ 3-\sqrt{6} \qquad \textbf{(D)}\ 2\sqrt{3}-3\qquad \textbf{(E)}\ \text{none of these}$

Solution

Using the difference to product identity, we find that $x=\cos 36^{\circ} - \cos 72^{\circ}$ is equivalent to \[x=\text{-}2\sin{\frac{(36^{\circ}+72^{\circ})}{2}}\sin{\frac{(36^{\circ}-72^{\circ})}{2}} \implies\] \[x=\text{-}2\sin54^{\circ}\sin(\text{-}18^{\circ}).\] Since sine is an odd function, we find that $\sin{(\text{-}18^{\circ})}= \text{-} \sin{18^{\circ}}$, and thus $\text{-}2\sin54^{\circ}\sin(\text{-}18^{\circ})=2\sin54^{\circ}\sin18^{\circ}$. Using the property $\sin{(90^{\circ}-a)}=\cos{a}$, we find \[x=2\cos(90^{\circ}-54^{\circ})\cos(90^{\circ}-18^{\circ}) \implies\] \[x=2\cos36^{\circ}\cos72^{\circ}.\] We multiply the entire expression by $\sin36^{\circ}$ and use the double angle identity of sine twice to find \[x\sin36^{\circ}=2\sin36^{\circ}\cos36^{\circ}\cos72^{\circ} \implies\] \[x\sin36^{\circ}=\sin72^{\circ}\cos72^{\circ} \implies\] \[x\sin36^{\circ}=\frac{1}{2}\sin144^{\circ}.\] Using the property $\sin(180^{\circ}-a)=\sin{a}$, we find $\sin144^{\circ}=\sin36^{\circ}.$ Substituting this back into the equation, we have \[x\sin36^{\circ}=\frac{1}{2}\sin36^{\circ}.\] Dividing both sides by $\sin36^{\circ}$, we have \[x=\boxed{\textbf{(B)}\ \frac{1}{2}}\]

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