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1975 AHSME Problems/Problem 30

Revision as of 07:16, 15 July 2024 by Saraliu24 (talk | contribs) (→‎Solution)

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 \[x=\boxed{\textbf{(B)}\ \frac{1}{2}}\]

Solution 2

We observe that $72^{\circ}$ is twice of $36^{\circ}$, so evaluating $\cos 36^{\circ}$ and using the double angle identify is enough to find $\cos 36^{\circ} - \cos 72^{\circ}$

We first let $n = \cos 36^{\circ}$. Notice that $36^{\circ}/4 = 9^{\circ}$ and $36^{\circ} = 45^{\circ} - 9^{\circ}$. Rewriting $\cos 36^{\circ}$ as $\cos (45^{\circ} - 9^{\circ})$, we then use the difference of angles identify to find \[n = \cos 36^{\circ} = \cos (45^{\circ} - 9^{\circ}) = \cos 45^{\circ}\cos 9^{\circ} - \sin 45^{\circ}\sin 9^{\circ} = \frac{\sqrt2}{2}(\cos 9^{\circ} - \sin 9^{\circ})\] Rearranged to get \[\sqrt2 n = \cos 9^{\circ} - \sin 9^{\circ}\] Using the half-angle identify twice, and noticing that $\cos 9^{\circ}$ is positive, we get \[\cos 9^{\circ} = \cos \frac{18^{\circ}}{2} = \sqrt \frac{1 + \cos 18^{\circ}}{2} = \sqrt \frac{1 + \sqrt \frac{1 + \cos 36^{\circ}}{2}}{2}\] Also, \[\sin 9^{\circ} = \sqrt {1 - \cos^2 9} = \sqrt {1 - \frac{1 + \sqrt{\frac{1 + 36^{\circ}}{2}}}{2}} = \sqrt {\frac{1 - \sqrt{\frac{1 + 36^{\circ}}{2}}}{2}}\]

Thus we can get \[\sqrt2 n = \sqrt \frac{1 + \sqrt \frac{1 + \cos 36^{\circ}}{2}}{2} - \sqrt {\frac{1 - \sqrt{\frac{1 + 36^{\circ}}{2}}}{2}}\] Squaring both sides and simplifying, this results in \[2n^2 = 1 - \sqrt{\frac{1-n}{2}}\] From which we get \[2n^2 - 1 = - \sqrt{\frac{1-n}{2}}\] Squaring both sides again, \[4n^4 - 4n^2 +1 = \frac{1-n}{2}\] Multiplying both sides by $2$ and rearranging, \[8n^4 - 8n^2 + n + 1 = 0\] Factoring out $8n^2$ from the first two terms and applying difference of squares, \[8n^2(n+1)(n-1) + n+1 = 0\] We know that $n = \cos 36^{\circ}$ which is not equal to $-1$, so we divide both sides by $n+1$ \[8n^3 -8n^2 +1 = 0\] Using the rational root test, we find that $n = \frac{1}{2}$ is a solution of this equation. However, we also know that $\cos 36^{\circ}$ is not equal to $\frac{1}{2}$, so we divide both sides by $n- \frac{1}{2}$ to get \[8n^2 - 4n - 2 = 0\] Using the quadratic formula, we find that \[n = \frac{1 \pm \sqrt{5}}{4}\] We know $n$ must be positive, which gives us \[n = \frac{1 + \sqrt{5}}{4}\]

Thus, using the double angle identify, we can find \[x = \cos 36^{\circ} - \cos 72^{\circ} = \cos 36^{\circ} - 2\cos^2 36^{\circ} + 1 = n - 2n^2 + 1 = \boxed{\textbf{(B)} \frac{1}{2}}\]

See Also

1975 AHSME (Problems • Answer Key • Resources)
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