Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 18"
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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> | <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 | + | as <math>x</math> varies over all numbers in the largest possible [[domain]] of <math>f</math>, is |
<center><math> \mathrm{(A) \ }-4 \qquad \mathrm{(B) \ }-2 \qquad \mathrm{(C) \ }0 \qquad \mathrm{(D) \ }2 \qquad \mathrm{(E) \ }4 </math></center> | <center><math> \mathrm{(A) \ }-4 \qquad \mathrm{(B) \ }-2 \qquad \mathrm{(C) \ }0 \qquad \mathrm{(D) \ }2 \qquad \mathrm{(E) \ }4 </math></center> |
Revision as of 15:41, 31 July 2006
Problem
The minimum value of the function
as varies over all numbers in the largest possible domain of , is
Solution
Recall the trigonometric identities
Since for real , we can now simplify the function to
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.