Triangular function and quadrilateral

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Triangular function and quadrilateral

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Source: National Defence Medical college Entrance exam November 2005

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Kunihiko_Chikaya

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Kunihiko_Chikaya

#1 h Jan 17, 2006, 4:29 PM • 2 Y

Y by Adventure10, Mango247

Given a convex quadrilateral such that $AB=AD=\sqrt{5},\ CB=CD=2, \tan \angle{DAB=-2,\ \angle{BCD}=2\alpha\ (0<2\alpha <\pi).}$ Find all the integers $n$ satisfying $\cos n\alpha=\cos \alpha.$

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#2 h Apr 29, 2025, 1:15 PM

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For given convex quadrilateral $ABCD$ with $AC \perp BD$ and $AC \cap BD = O \Rightarrow OB=OD =x$ . WLOG, right triangles $\Delta OAB$ and $\Delta OBC$ have the equations:

$\sin\left[\frac{\arctan(-2)}{2}\right] = \frac{x}{\sqrt{5}}$ (i);

$\sin(\alpha) = \frac{x}{2}$ (ii)

of which equating (i) with (ii) yields:

$2\sin(\alpha) = \sqrt{5}\sin\left[\frac{\arctan(-2)}{2}\right]$ ;

or $2\sin(\alpha) = \sqrt{5}\sqrt{\frac{1-\cos[\arccos(-1/\sqrt{5})]}{2}}$ ;

or $\sin(\alpha) = \sqrt{\frac{5+\sqrt{5}}{8}} = \sqrt{\frac{1}{2}\left(1+\frac{1+\sqrt{5}}{4}\right)} = \sqrt{\frac{1+\cos(\pi/5)}{2}} = \cos\left(\frac{\pi}{10}\right) = \sin\left(\frac{\pi}{2}-\frac{\pi}{10}\right)$ ;

or $\alpha = \frac{2\pi}{5}$ .

We now wish to determine all $n \in \mathbb{Z}$ such that $\cos(n\alpha) = \cos(\alpha)$ . Knowing that $\cos(\alpha) > 0$ in the 1st and 4th Quadrants of the $xy-$ plane, and that $5k\alpha = 2k\pi \Rightarrow$ we require $\textcolor{red}{n = 5k \pm 1}$ for all $k \in \mathbb{Z}$ .

This post has been edited 8 times. Last edited by Mathzeus1024, Apr 29, 2025, 2:40 PM

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