Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 10"

 
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== Problem ==
 
== Problem ==
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<math>\arcsin\left(\frac{1}{3}\right) + \arccos\left(\frac{1}{3}\right) + \arctan\left(\frac13\right) + \text{arccot}\left(\frac13\right) =</math>
  
<center><math> \mathrm{(A) \ } \qquad \mathrm{(B) \ } \qquad \mathrm{(C) \ } \qquad \mathrm{(D) \ } \qquad \mathrm{(E) \ }  </math></center>
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<center><math> \mathrm{(A) \ }\pi \qquad \mathrm{(B) \ }\pi/2 \qquad \mathrm{(C) \ }\pi/3 \qquad \mathrm{(D) \ }2\pi/3 \qquad \mathrm{(E) \ }3/\pi/4 </math></center>
  
 
== Solution ==
 
== Solution ==
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If we construct right triangles for each pair of arguments (<math>\arcsin, \arccos</math> in one triangle and <math>\arctan, \text{arccot}</math> in another), we see that the sum of the angles is <math>90^\circ+90^\circ=\pi</math>.
  
== See also ==
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* [[University of South Carolina High School Math Contest/1993 Exam]]
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* [[University of South Carolina High School Math Contest/1993 Exam/Problem 9|Previous Problem]]
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* [[University of South Carolina High School Math Contest/1993 Exam/Problem 11|Next Problem]]
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* [[University of South Carolina High School Math Contest/1993 Exam|Back to Exam]]
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[[Category:Introductory Trigonometry Problems]]

Latest revision as of 21:15, 31 May 2009

Problem

$\arcsin\left(\frac{1}{3}\right) + \arccos\left(\frac{1}{3}\right) + \arctan\left(\frac13\right) + \text{arccot}\left(\frac13\right) =$

$\mathrm{(A) \ }\pi \qquad \mathrm{(B) \ }\pi/2 \qquad \mathrm{(C) \ }\pi/3 \qquad \mathrm{(D) \ }2\pi/3 \qquad \mathrm{(E) \ }3/\pi/4$

Solution

If we construct right triangles for each pair of arguments ($\arcsin, \arccos$ in one triangle and $\arctan, \text{arccot}$ in another), we see that the sum of the angles is $90^\circ+90^\circ=\pi$.