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

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== Problem ==
 
== Problem ==
<math>\arcsin\left(\frac{1}{3}\right) + \arccos\left(\frac{1}{3}\right) + \arctan\left(\frac13\right) + \arccot\left(\frac13\right) =</math>
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<math>\arcsin\left(\frac{1}{3}\right) + \arccos\left(\frac{1}{3}\right) + \arctan\left(\frac13\right) + arccot\left(\frac13\right) =</math>
  
 
<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>
 
<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>

Revision as of 17:12, 11 October 2007

Problem

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

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

Solution

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