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> | + | <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) \ }\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> | ||
== Solution == | == Solution == | ||
| − | If we construct right triangles for each pair of arguments (<math>\arcsin, \arccos</math> in one triangle and <math>\arctan, \arccot</math> in another), we see that the sum of the angles is <math>90^\circ+90^\circ=\pi</math>. | + | 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>. |
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Latest revision as of 21:15, 31 May 2009
Problem
Solution
If we construct right triangles for each pair of arguments (
in one triangle and 
in another), we see that the sum of the angles is 
.
