Answered: "Showing that 1−sin⁡5π183sin⁡5π18=tan⁡π18"

deiteresfp

Answered question

2022-01-16

Showing that

\frac{1 - \sin \frac{5 π}{18}}{\sqrt{3} \sin \frac{5 π}{18}} = \tan \frac{π}{18}

Jeffery Autrey

Beginner2022-01-17Added 35 answers

Evaluate,

\frac{1}{\sin \frac{5 π}{18}} - \sqrt{3} \tan \frac{π}{18} = \frac{\cos \frac{π}{18} - \sqrt{3} \sin \frac{5 π}{18} \sin \frac{π}{18}}{\sin \frac{5 π}{18} \cos \frac{π}{18}}

(1)
Examine the numerator,

\cos \frac{π}{18} - \frac{\sqrt{3}}{2} \cdot 2 \sin \frac{5 π}{18} \sin \frac{π}{18}

= \cos \frac{π}{18} - \cos \frac{π}{6} (\cos \frac{2 π}{9} - \cos \frac{π}{3})

= \cos \frac{π}{18} - \frac{12}{\cos \frac{π}{18} + \cos \frac{7 π}{18}} + \cos \frac{π}{6} \cos \frac{π}{3}

= \frac{12}{\cos \frac{π}{18} - \cos \frac{7 π}{18}} + \frac{12}{\cos \frac{π}{6}}

= \sin \frac{π}{6} \sin \frac{2 π}{9} + \frac{12}{\cos \frac{π}{6}}

= \frac{12}{\sin \frac{2 π}{9} + \sin \frac{π}{3}} = \sin \frac{5 π}{18} \cos \frac{π}{18}

Substitute the result for the numerator into (1) to have

\frac{1}{\sin \frac{5 π}{18}} - \sqrt{3} \tan \frac{π}{18} = 1

Then, rearrange to obtain,

\frac{1 - \sin \frac{5 π}{18}}{\sqrt{3} \sin \frac{5 π}{18}} = \tan \frac{π}{18}

RizerMix

Expert2022-01-20Added 656 answers

We need to prove that

(1 - \sin 50^{\circ}) \cos 10^{\circ} = 2 \sin 50^{\circ} \sin 10^{\circ} \cos 30^{\circ}

2 \cos 10^{\circ} - \sin 60^{\circ} - \sin 40^{\circ} = 2 (\cos 40^{\circ} - \cos 60^{\circ}) \cos 30^{\circ}

2 \cos 10^{\circ} - \sin 40^{\circ} = 2 \cos 40^{\circ} \cos 30^{\circ}

2 \cos 10^{\circ} - \sin 40^{\circ} = \cos 70^{\circ} + \cos 10^{\circ}

\cos 10^{\circ} = \cos 70^{\circ} + \cos 50^{\circ}

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