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Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 18"

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Recall the [[trigonometric identities]]
 
Recall the [[trigonometric identities]]
  
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<center><math> \sin^2 x + \cos^2 x = 1 </math> </center>
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{| class="wikitable" style="margin: 1em auto 1em auto"
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<center><math> \tan^2 x + 1 = \sec^2 x </math> </center>
+
| <center><math> \sin^2 x + \cos^2 x = 1 </math> </center>
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<center><math> 1 + \cot^2 x = \csc^2 x </math> </center>
+
|-
 +
| <center><math> \tan^2 x + 1 = \sec^2 x </math> </center>
 +
|-
 +
| <center><math> 1 + \cot^2 x = \csc^2 x </math> </center>
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|}
  
 
We can now simplify the function to
 
We can now simplify the function to

Revision as of 14:50, 31 July 2006

Problem

The minimum value of the function

$\displaystyle f(x) = \frac{\sin (x)}{\sqrt{1 - \cos^2 (x)}} + \frac{\cos(x)}{\sqrt{1 - \sin^2 (x) }} + \frac{\tan(x)}{\sqrt{\sec^2 (x) - 1}} + \frac{\cot (x)}{\sqrt{\csc^2 (x) - 1}}$

as $x$ varies over all numbers in the largest possible domain of $f$, is

$\mathrm{(A) \ }-4 \qquad \mathrm{(B) \ }-2 \qquad \mathrm{(C) \ }0 \qquad \mathrm{(D) \ }2 \qquad \mathrm{(E) \ }4$

Solution

Recall the trigonometric identities

$\sin^2 x + \cos^2 x = 1$
$\tan^2 x + 1 = \sec^2 x$
$1 + \cot^2 x = \csc^2 x$

We can now simplify the function to

$f(x) = \frac{\sin(x)}{|\sin (x)|}+\frac{\cos(x)}{|\cos(x)|} + \frac{\tan(x)}{|\tan(x)|} + \frac{\cot(x)}{|\cot(x)|}.$

Now we must consider the quadrant that $x$ is in. If $x$ is in quadrant I, then all of the trig functions are positive and $f(x)=1+1+1+1=4$. If $x$ is in quadrant II, then sine is positive and the rest of cosine, tangent, and cotangent are negative giving $f(x)=1-1-1-1=-2$. If $x$ is in quadrant III, then tangent and cotangent are positive while sine and cosine are negative making $f(x)=1+1-1-1=0$. Finally, if $x$ is in quadrant IV, then only cosine is positive with the other three being negative giving $f(x)=-1+1-1-1=-2$. Thus our answer is -2.

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