Prove Existence of Solution: POTW #290 Nov 26, 2017

[anemone]

Here is this week's POTW:

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Prove that there exists $x\in \Bbb{N}$, where $1\le x \le 89$ such that $\sqrt{3}\tan x^\circ-1=\sec 20^\circ$.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Congratulations to the following members for their correct solution: (Smile)

1. Opalg
2. kaliprasad
3. greg1313

Solution from Opalg:

Spoiler

A quick check on a calculator suggests that the answer should be $x=50$. With that in mind, use the addition formulas for sin and cos, and the values of $\sin 30^\circ$ and $\cos 30^\circ$, to get $$\tan 50^\circ = \frac{\sin(30^\circ+20^\circ)} {\cos(30^\circ+20^\circ)} = \frac{\frac12(\cos 20^\circ + \sqrt3\sin 20^\circ)}{\frac12(\sqrt3\cos 20^\circ - \sin 20^\circ)},$$ $$ \sqrt3\tan 50^\circ - 1 = \frac{\sqrt3(\cos 20^\circ + \sqrt3\sin 20^\circ) - (\sqrt3\cos 20^\circ - \sin 20^\circ)}{\sqrt3\cos 20^\circ - \sin 20^\circ} = \frac{4\sin 20^\circ }{\sqrt3\cos 20^\circ - \sin 20^\circ}.\qquad(1)$$ The denominator in (1) is $$\sqrt3\cos 20^\circ - \sin 20^\circ = 2\cos 50^\circ = 2\sin40^\circ = 4\sin20^\circ\cos20^\circ.$$ It follows from (1) that $ \sqrt3\tan 50^\circ - 1 = \dfrac{4\sin 20^\circ }{4\sin20^\circ\cos20^\circ} = \dfrac1{\cos20^\circ} = \sec20^\circ$ , as required.