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Difference between revisions of "1984 AHSME Problems/Problem 15"

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NOTE: To show that $ 18^\circ $ is not the only solution, we can also set $ 5x $ equal to another angle measure congruent to $ \frac{\pi}{2} $, such as $ \frac{5\pi}{2} $, yielding another solution as $ \frac{\pi}{2}=90^\circ $, which clearly is a solution to the equation.
 
NOTE: To show that $ 18^\circ $ is not the only solution, we can also set $ 5x $ equal to another angle measure congruent to $ \frac{\pi}{2} $, such as $ \frac{5\pi}{2} $, yielding another solution as $ \frac{\pi}{2}=90^\circ $, which clearly is a solution to the equation.
 
==Solution 2==
 
==Solution 2==
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We can simply try the answers. We quickly see that $ 18^\circ$ works, since $sin(36^\circ)sin(54^\circ)=cos(54^\circ)cos(36^\circ)=cos(36^\circ)cos(54^\circ)$ by the identity $cos(x)=sin(90^\circ-c)$
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We can simply try the answers. We quickly see that <math> 18^\circ</math> works, since <math>\sin(36^\circ)\sin(54^\circ)=\cos(54^\circ)\cos(36^\circ)=\cos(36^\circ)\cos(54^\circ)</math> by the identity <math>\cos(x)=\sin(90^\circ-x)</math>
  
 
==See Also==
 
==See Also==
 
{{AHSME box|year=1984|num-b=14|num-a=16}}
 
{{AHSME box|year=1984|num-b=14|num-a=16}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 15:51, 1 March 2015

Problem 15

If $\sin{2x}\sin{3x}=\cos{2x}\cos{3x}$, then one value for $x$ is

$\mathrm{(A) \ }18^\circ \qquad \mathrm{(B) \ }30^\circ \qquad \mathrm{(C) \ } 36^\circ \qquad \mathrm{(D) \ }45^\circ \qquad \mathrm{(E) \ } 60^\circ$

Solution

We divide both sides of the equation by $\cos{2x}\times\cos{3x}$ to get $\frac{\sin{2x}\times\sin{3x}}{\cos{2x}\times\cos{3x}}=1$, or $\tan{2x}\times\tan{3x}=1$.

This looks a lot like the formula relating the slopes of two perpendicular lines, which is $m_1\times m_2=-1$, where $m_1$ and $m_2$ are the slopes. It's made even more relatable by the fact that the tangent of an angle can be defined by the slope of the line that makes that angle with the x-axis.

We can make this look even more like the slope formula by multiplying both sides by $-1$:

$\tan{2x}\times-\tan{3x}=-1$, and using the trigonometric identity $-\tan{x}=\tan{-x}$, we have $\tan{2x}\times\tan{-3x}=-1$.

Now it's time for a diagram:

[asy] unitsize(2.54cm); draw(unitcircle); draw((0,-1.25)--(0,1.25)); draw((-1.25,0)--(1.25,0)); draw((0,0)--(cos(2pi/10),sin(2pi/10))); draw((0,0)--(cos(-3pi/10),sin(-3pi/10))); label("$\tan{-3x}$",(cos(-3pi/10),sin(-3pi/10)),SE); label("$\tan{2x}$",(cos(2pi/10),sin(2pi/10)),NE); label("$2x$",(.125,.03),ENE); label("$3x$",(.125,-.06),ESE); [/asy] Since the product of the two slopes, $ \tan{2x} $ and $ \tan{-3x} $, is $ -1 $, the lines are perpendicular, and the angle between them is $ \frac{\pi}{2} $. The angle between them is also $ 2x+3x=5x $, so $ 5x=\frac{\pi}{2} $ and $ x=\frac{\pi}{10} $, or $ 18^\circ, \boxed{\text{A}} $.

NOTE: To show that $ 18^\circ $ is not the only solution, we can also set $ 5x $ equal to another angle measure congruent to $ \frac{\pi}{2} $, such as $ \frac{5\pi}{2} $, yielding another solution as $ \frac{\pi}{2}=90^\circ $, which clearly is a solution to the equation.

Solution 2

We can simply try the answers. We quickly see that $18^\circ$ works, since $\sin(36^\circ)\sin(54^\circ)=\cos(54^\circ)\cos(36^\circ)=\cos(36^\circ)\cos(54^\circ)$ by the identity $\cos(x)=\sin(90^\circ-x)$

See Also

1984 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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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