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> | ||
== Solution == | == Solution == | ||
− | Recall the [[ | + | Recall the [[trigonometric identities]] |
− | <center><math> \sin^2 x + \cos^2 x = 1 </math> </center> | + | {| class="wikitable" style="margin: 1em auto 1em auto" |
− | <center><math> \tan^2 x + 1 = \sec^2 x </math> </center> | + | | <center><math> \sin^2 x + \cos^2 x = 1 </math> </center> |
− | <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> | ||
+ | |} | ||
− | + | 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> | ||
− | Now we must consider the quadrant that <math>x</math> is in. If <math>x</math> is in quadrant I, then all of the trig functions are positive and <math>f(x)=1+1+1+1=4</math>. If <math>x</math> is in quadrant II, then sine is positive and the rest of cosine, tangent, and cotangent are negative giving <math>f(x)=1-1-1-1=-2</math>. If <math>x</math> is in quadrant III, then tangent and cotangent are positive while sine and cosine are negative making <math>f(x)=1+1-1-1=0</math>. Finally, if <math>x</math> is in quadrant IV, then only cosine is positive with the other three being negative giving <math>f(x)=-1+1-1-1=-2</math>. Thus our answer is -2. | + | Now we must consider the [[quadrant]] that <math>x</math> is in. If <math>x</math> is in quadrant I, then all of the trig functions are positive and <math>f(x)=1+1+1+1=4</math>. If <math>x</math> is in quadrant II, then sine is positive and the rest of cosine, tangent, and cotangent are negative giving <math>f(x)=1-1-1-1=-2</math>. If <math>x</math> is in quadrant III, then tangent and cotangent are positive while sine and cosine are negative, making <math>f(x)=1+1-1-1=0</math>. Finally, if <math>x</math> is in quadrant IV, then only cosine is positive with the other three being negative giving <math>f(x)=-1+1-1-1=-2</math>. Thus our answer is -2. |
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Latest revision as of 00:37, 26 January 2023
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.