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University of South Carolina High School Math Contest/1993 Exam/Problem 28

Revision as of 12:37, 29 August 2006 by Bictor717 (talk | contribs)

Problem

Suppose $\triangle ABC$ is a triangle with 3 acute angles $A, B,$ and $C$. Then the point $( \cos B - \sin A, \sin B - \cos A)$

(A) can be in the 1st quadrant and can be in the 2nd quadrant only

(B) can be in the 3rd quadrant and can be in the 4th quadrant only

(C) can be in the 2nd quadrant and can be in the 3rd quadrant only

(D) can be in the 2nd quadrant only

(E) can be in any of the 4 quadrants

Solution

This problem needs a solution. If you have a solution for it, please help us out by adding it.

First note that $\cos{A}, \cos{B}, \sin{A}, \sin{B}>0$ and $A+B>90$.

Since $\cos B+\sin A>0$, $\cos B-\sin A$ and $\cos^2B-\sin^2A$ have the same sign. Similarly, $\sin B-\cos A$ and $\sin^2B-\cos^2A$ have the same sign.

Notice that $\cos^2B-\sin^2A=-(\sin^2B-\cos^2A)$. Then $(\cos B-\sin A, \sin B-\cos A)$ can only lie in the 2nd and 4th quadrants.

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