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

(Created page with "==Problem 30== Let <math>x=\cos 36^{\circ} - \cos 72^{\circ}</math>. Then <math>x</math> equals <math>\textbf{(A)}\ \frac{1}{3}\qquad \textbf{(B)}\ \frac{1}{2} \qquad \textbf...")
 
(→‎Solution)
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==Solution==
 
==Solution==
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Using the difference to product identity, we find that
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<math>x=\cos 36^{\circ} - \cos 72^{\circ}</math> is equivalent to <cmath>x=-2\sin{\frac{(36^{\circ}+72^{\circ})}{2}}\sin{\frac{(36^{\circ}-72^{\circ})}{2}} \implies</cmath>
 +
<cmath>x=-2\sin54^{\circ}\sin(\text{-}18^{\circ}).</cmath>
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Since sine is an odd function, we find that <math>\sin{(\text{-}18^{\circ})}= - \sin{18^{\circ}}</math>, and thus <math>-2\sin54^{\circ}\sin(\text{-}18^{\circ})=2\sin54^{\circ}\sin18^{\circ}</math>. Using the property <math>\sin{(90^{\circ}-a)}=\cos{a}</math>, we find
 +
<cmath>x=2\cos(90^{\circ}-54^{\circ})\cos(90^{\circ}-18^{\circ}) \implies</cmath>
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<cmath>x=2\cos36^{\circ}\cos72^{\circ}.</cmath>
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We multiply the entire expression by <math>\sin36^{\circ}</math> and use the double angle identity of sine twice to find
 +
<cmath>x\sin36^{\circ}=2\sin36^{\circ}\cos36^{\circ}\cos72^{\circ} \implies</cmath>
 +
<cmath>x\sin36^{\circ}=\sin72^{\circ}\cos72^{\circ} \implies</cmath>
 +
<cmath>x\sin36^{\circ}=\frac{1}{2}\sin144^{\circ}.</cmath>
 +
Using the property <math>\sin(180^{\circ}-a)=\sin{a}</math>, we find <math>\sin144^{\circ}=\sin36^{\circ}.</math> Substituting this back into the equation, we have
 +
<cmath>x\sin36^{\circ}=\frac{1}{2}\sin36^{\circ}.</cmath>
 +
Dividing both sides by <math>\sin36^{\circ}</math>, we have
 +
<cmath>x=\boxed{\textbf{(B)}\ \frac{1}{2}}</cmath>

Revision as of 16:29, 14 July 2016

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 \[x=-2\sin{\frac{(36^{\circ}+72^{\circ})}{2}}\sin{\frac{(36^{\circ}-72^{\circ})}{2}} \implies\] \[x=-2\sin54^{\circ}\sin(\text{-}18^{\circ}).\] Since sine is an odd function, we find that $\sin{(\text{-}18^{\circ})}= - \sin{18^{\circ}}$, and thus $-2\sin54^{\circ}\sin(\text{-}18^{\circ})=2\sin54^{\circ}\sin18^{\circ}$. 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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