Difference between revisions of "1996 AIME Problems/Problem 10"

Revision as of 18:33, 4 July 2013

Problem

Find the smallest positive integer solution to $\tan{19x^{\circ}}=\dfrac{\cos{96^{\circ}}+\sin{96^{\circ}}}{\cos{96^{\circ}}-\sin{96^{\circ}}}$ .

Solution

$\dfrac{\cos{96^{\circ}}+\sin{96^{\circ}}}{\cos{96^{\circ}}-\sin{96^{\circ}}} =$ $\dfrac{\sin{186^{\circ}}+\sin{96^{\circ}}}{\sin{186^{\circ}}-\sin{96^{\circ}}} =$ $\dfrac{\sin{(141^{\circ}+45^{\circ})}+\sin{(141^{\circ}-45^{\circ})}}{\sin{(141^{\circ}+45^{\circ})}-\sin{(141^{\circ}-45^{\circ})}} =$ $\dfrac{2\sin{141^{\circ}}\cos{45^{\circ}}}{2\cos{141^{\circ}}\sin{45^{\circ}}} = \tan{141^{\circ}}$ .

The period of the tangent function is $180^\circ$ , and the tangent function is one-to-one over each period of its domain.

Thus, $19x \equiv 141 \pmod{180}$ .

Since $19^2 \equiv 361 \equiv 1 \pmod{180}$ , multiplying both sides by $19$ yields $x \equiv 141 \cdot 19 \equiv (140+1)(18+1) \equiv 0+140+18+1 \equiv 159 \pmod{180}$ .

Therefore, the smallest positive solution is $x = \boxed{159}$ .

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	[[Category:Intermediate Trigonometry Problems]]		[[Category:Intermediate Trigonometry Problems]]
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Difference between revisions of "1996 AIME Problems/Problem 10"

Revision as of 18:33, 4 July 2013

Problem

Solution

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