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 "1974 AHSME Problems/Problem 22"

(Created page with "==Problem== The minimum of <math> \sin\frac{A}{2}-\sqrt{3}\cos\frac{A}{2} </math> is attained when <math> A </math> is <math> \mathrm{(A)\ } -180^\circ \qquad \mathrm{(B) \ }60...")
 
 
(One intermediate revision by one other user not shown)
Line 11: Line 11:
 
==See Also==
 
==See Also==
 
{{AHSME box|year=1974|num-b=21|num-a=23}}
 
{{AHSME box|year=1974|num-b=21|num-a=23}}
 +
[[Category:Introductory Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 11:43, 5 July 2013

Problem

The minimum of $\sin\frac{A}{2}-\sqrt{3}\cos\frac{A}{2}$ is attained when $A$ is

$\mathrm{(A)\ } -180^\circ \qquad \mathrm{(B) \ }60^\circ \qquad \mathrm{(C) \ } 120^\circ \qquad \mathrm{(D) \ } 0^\circ \qquad \mathrm{(E) \ }\text{none of these}$

Solution

Define a new number $B$ such that $\cos(B)=\frac{1}{2}$ and $\sin(B)=-\frac{\sqrt{3}}{2}$. Notice that $B=\frac{5\pi}{3}$. Therefore, we have $\sin\frac{A}{2}-\sqrt{3}\cos\frac{A}{2}= 2\cos(B)\sin\left(\frac{A}{2}\right)+2\sin(B)\cos\left(\frac{A}{2}\right)$.

Recognizing the angle sum formula, we can rewrite this as $\sin\frac{A}{2}-\sqrt{3}\cos\frac{A}{2}=2\sin\left(\frac{A}{2}+B\right)$. We can now clearly see that the minimum is $-2$, and this is achieved when $\frac{A}{2}+B=\frac{3\pi}{2}$. Plugging in $B=\frac{5\pi}{3}$, we get $\frac{A}{2}=-\frac{\pi}{6}\implies A=-\frac{\pi}{3}=-60^\circ$. $\boxed{\text{E}}$

See Also

1974 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
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=1974_AHSME_Problems/Problem_22&oldid=56274"