Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 18"
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− | + | Since <math>\sqrt{x^2} = |x|</math> for [[real number|real]] <math>x</math>, we can now simplify the [[function]] to | |
<center><math> f(x) = \frac{\sin(x)}{|\sin (x)|}+\frac{\cos(x)}{|\cos(x)|} + \frac{\tan(x)}{|\tan(x)|} + \frac{\cot(x)}{|\cot(x)|}. </math></center> | <center><math> f(x) = \frac{\sin(x)}{|\sin (x)|}+\frac{\cos(x)}{|\cos(x)|} + \frac{\tan(x)}{|\tan(x)|} + \frac{\cot(x)}{|\cot(x)|}. </math></center> |
Revision as of 15:40, 31 July 2006
Problem
The minimum value of the function
![$\displaystyle f(x) = \frac{\sin (x)}{\sqrt{1 - \cos^2 (x)}} + \frac{\cos(x)}{\sqrt{1 - \sin^2 (x) }} + \frac{\tan(x)}{\sqrt{\sec^2 (x) - 1}} + \frac{\cot (x)}{\sqrt{\csc^2 (x) - 1}}$](http://latex.artofproblemsolving.com/c/6/a/c6a4e021d142f4bad5a913725ac6e8f1316b9066.png)
as varies over all numbers in the largest possible domain of
, is
![$\mathrm{(A) \ }-4 \qquad \mathrm{(B) \ }-2 \qquad \mathrm{(C) \ }0 \qquad \mathrm{(D) \ }2 \qquad \mathrm{(E) \ }4$](http://latex.artofproblemsolving.com/2/f/7/2f7612b8e989eded4129ff00a31811dee6a6ef5e.png)
Solution
Recall the trigonometric identities
![]() |
![]() |
![]() |
Since for real
, we can now simplify the function to
![$f(x) = \frac{\sin(x)}{|\sin (x)|}+\frac{\cos(x)}{|\cos(x)|} + \frac{\tan(x)}{|\tan(x)|} + \frac{\cot(x)}{|\cot(x)|}.$](http://latex.artofproblemsolving.com/4/3/a/43a5bd11062d3c08c0c1ffb47bf73377607b04e0.png)
Now we must consider the quadrant that is in. If
is in quadrant I, then all of the trig functions are positive and
. If
is in quadrant II, then sine is positive and the rest of cosine, tangent, and cotangent are negative giving
. If
is in quadrant III, then tangent and cotangent are positive while sine and cosine are negative making
. Finally, if
is in quadrant IV, then only cosine is positive with the other three being negative giving
. Thus our answer is -2.