New Year's Countdown Day 20: Sine ± \pm Cosine

Geometry Level 3

Let x x be a real number such that

tan ( 20 x ) = sin ( 1 7 ) + cos ( 1 7 ) sin ( 1 7 ) cos ( 1 7 ) . \tan(20x^{\circ}) = \dfrac{\sin(17^{\circ})+ \cos(17^{\circ})}{\sin(17^{\circ}) - \cos(17^{\circ})}.

The minimum possible value of x |x| can be written in the form p q , \dfrac{p}{q}, where p p and q q are coprime positive integers. Find the value of p + q . p + q.

Notation: |\cdot| denotes the absolute value function .


This problem is part of the set New Year's Countdown 2017 .


The answer is 41.

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1 solution

Chew-Seong Cheong
Dec 21, 2017

tan ( 20 x ) = sin 1 7 + cos 1 7 sin 1 7 cos 1 7 = 2 ( 1 2 sin 1 7 + 1 2 cos 1 7 ) 2 ( 1 2 cos 1 7 1 2 sin 1 7 ) = sin 1 7 cos 4 5 + cos 1 7 sin 4 5 cos 1 7 cos 4 5 sin 1 7 sin 4 5 = sin 6 2 cos 6 2 = tan 6 2 \begin{aligned} \tan(20x^\circ) = \frac {\sin 17^\circ + \cos 17^\circ}{\sin 17^\circ - \cos 17^\circ} & = \frac {\sqrt 2 \left(\frac 1{\sqrt 2} \sin 17^\circ + \frac 1{\sqrt 2} \cos 17^\circ\right)}{-\sqrt 2 \left(\frac 1{\sqrt 2} \cos 17^\circ - \frac 1{\sqrt 2} \sin 17^\circ\right)} = - \frac {\sin 17^\circ \cos 45^\circ +\cos 17^\circ \sin 45^\circ}{\cos 17^\circ \cos 45^\circ - \sin 17^\circ \sin 45^\circ} = - \frac {\sin 62^\circ}{\cos 62^\circ} = -\tan 62^\circ \end{aligned}

The smallest x |x| comes from 20 x = 6 2 x = 3.1 = 31 10 20x = - 62^\circ\implies |x| = 3.1 = \dfrac {31}{10} p + q = 31 + 10 = 41 \implies p+q = 31+10 = \boxed{41} .

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