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Difference between revisions of "1975 AHSME Problems/Problem 30"

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Dividing both sides by <math>\sin36^{\circ}</math>, we have
 
Dividing both sides by <math>\sin36^{\circ}</math>, we have
 
<cmath>x=\boxed{\textbf{(B)}\ \frac{1}{2}}</cmath>
 
<cmath>x=\boxed{\textbf{(B)}\ \frac{1}{2}}</cmath>
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 +
==Solution 2==
 +
 +
We observe that <math>72^{\circ}</math> is twice of <math>36^{\circ}</math>, so evaluating <math>\cos 36^{\circ}</math> and using the double angle identify is enough to find <math>\cos 36^{\circ} - \cos 72^{\circ}</math>
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 +
We first let <math>n = \cos 36^{\circ}</math>.
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Notice that <math>36^{\circ}/4 = 9^{\circ}</math> and <math>36^{\circ} = 45^{\circ} - 9^{\circ}</math>.
 +
Rewriting <math>\cos 36^{\circ}</math> as <math>\cos (45^{\circ} - 9^{\circ})</math>, we then use the difference of angles identify to find
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<cmath>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})</cmath>
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Rearranged to get
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<cmath>\sqrt2 n = \cos 9^{\circ} - \sin 9^{\circ}</cmath>
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Using the half-angle identify twice, and noticing that <math>\cos 9^{\circ}</math> is positive, we get
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<cmath> \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}</cmath>
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Also,
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<cmath>\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}}</cmath>
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 +
Thus we can get
 +
<cmath> \sqrt2 n = \sqrt \frac{1 + \sqrt \frac{1 + \cos 36^{\circ}}{2}}{2} - \sqrt {\frac{1 - \sqrt{\frac{1 + 36^{\circ}}{2}}}{2}}</cmath>
 +
Squaring both sides and simplifying, this results in
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<cmath>2n^2 = 1 - \sqrt{\frac{1-n}{2}}</cmath>
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From which we get
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<cmath> 2n^2 - 1 = - \sqrt{\frac{1-n}{2}}</cmath>
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Squaring both sides again,
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<cmath> 4n^4 - 4n^2 +1 = \frac{1-n}{2}</cmath>
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Multiplying both sides by <math>2</math> and rearranging,
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<cmath>8n^4 - 8n^2 + n + 1 = 0</cmath>
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Factoring out <math>8n^2</math> from the first two terms and applying difference of squares,
 +
<cmath>8n^2(n+1)(n-1) + n+1 = 0</cmath>
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We know that <math>n = \cos 36^{\circ}</math> which is not equal to <math>-1</math>, so we divide both sides by <math>n+1</math>
 +
<cmath> 8n^3 -8n^2 +1 = 0</cmath>
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Using the rational root test, we find that <math>n = \frac{1}{2}</math> is a solution of this equation. However, we also know that <math>\cos 36^{\circ}</math> is not equal to <math>\frac{1}{2}</math>, so we divide both sides by <math>n- \frac{1}{2}</math> to get
 +
<cmath> 8n^2 - 4n - 2 = 0</cmath>
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Using the quadratic formula, we find that
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<cmath>n = \frac{1 \pm \sqrt{5}}{4}</cmath>
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We know <math>n</math> must be positive, which gives us
 +
<cmath>n = \frac{1 + \sqrt{5}}{4}</cmath>
 +
 +
Thus, using the double angle identify, we can find
 +
<cmath>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}} </cmath>
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 +
-SaraLiu24
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 +
==See Also==
 +
{{AHSME box|year=1975|num-b=28|after=Last Problem}}
 +
{{MAA Notice}}

Latest revision as of 22:11, 15 July 2024

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 \[x = \cos 36^{\circ} - \cos 72^{\circ} = \cos 36^{\circ} - 2\cos^2 36^{\circ} + 1 = n - 2n^2 + 1 = \boxed{\textbf{(B)} \frac{1}{2}}\]

-SaraLiu24

See Also

1975 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 28 		Followed by
Last Problem
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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