Difference between revisions of "1974 AHSME Problems/Problem 22"

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 $\frac{A}{2}=-\frac{\pi}{6}\implies A=-\frac{\pi}{3}=-60^\circ$ . $\boxed{\text{E}}$

1974 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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All AHSME Problems and Solutions

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

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 {{AHSME box|year=1974|num-b=21|num-a=23}}
 [[Category:Introductory Algebra Problems]]
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Difference between revisions of "1974 AHSME Problems/Problem 22"

Latest revision as of 11:43, 5 July 2013

Problem

Solution

See Also