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

 
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== Problem ==
 
== Problem ==
 +
The minimum value of the function
  
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<center><math> \mathrm{(A) \ } \qquad \mathrm{(B) \ } \qquad \mathrm{(C) \ } \qquad \mathrm{(D) \ } \qquad \mathrm{(E) \ }  </math></center>
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<center><math>\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}}</math></center>
 +
 
 +
as <math>x</math> varies over all numbers in the largest possible domain of <math>f</math>, is
 +
 
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<center><math> \mathrm{(A) \ }-4 \qquad \mathrm{(B) \ }-2 \qquad \mathrm{(C) \ }0 \qquad \mathrm{(D) \ }2 \qquad \mathrm{(E) \ }4 </math></center>
  
 
== Solution ==
 
== Solution ==
 +
Recall the [[Pythagorean Identities]]:
 +
 +
<center><math> \sin^2 x + \cos^2 x = 1 </math> </center>
 +
<center><math> \tan^2 x + 1 = \sec^2 x </math> </center>
 +
<center><math> 1 + \cot^2 x = \csc^2 x </math> </center>
 +
 +
We can now simplify the function to
 +
 +
<center><math> f(x) = \frac{\sin(x)}{\pm \sin (x)}+\frac{\cos(x)}{\pm \cos(x)} + \frac{\tan(x)}{\pm \tan(x)} + \frac{\cot(x)}{\pm \cot(x)} </math></center>
 +
 +
which is just <math>\pm 1 \pm 1 \pm 1 \pm 1</math>.  The minimum value is thus -4.
  
 
== See also ==
 
== See also ==
 
* [[University of South Carolina High School Math Contest/1993 Exam]]
 
* [[University of South Carolina High School Math Contest/1993 Exam]]

Revision as of 19:10, 22 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 Pythagorean 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

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

which is just $\pm 1 \pm 1 \pm 1 \pm 1$. The minimum value is thus -4.

See also

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