Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution)
(→‎Solution)
Line 19: Line 19:
 
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>
 +
 +
==See Also==
 +
{{AHSME box|year=1975|num-b=28|Last Problem}}
 +
{{MAA Notice}}

Revision as of 21:45, 15 May 2022

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

See Also

1975 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 28 		Followed by
[[1975 AHSME Problems/Problem {{{num-a}}}|Problem {{{num-a}}}]]
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1975_AHSME_Problems/Problem_30&oldid=173985"