Check the solution to "I found this interesting result. Show that1+sin⁡6∘+cos⁡12∘cos⁡6∘+sin⁡12∘=3"

Liseskirlsojh

Answered question

2022-03-30

I found this interesting result. Show that

\frac{1 + {\sin 6}^{\circ} + {\cos 12}^{\circ}}{{\cos 6}^{\circ} + {\sin 12}^{\circ}} = \sqrt{3}

kattylouxlvc

Beginner2022-03-31Added 11 answers

1 + {\sin 6}^{\circ} + {\cos 12}^{\circ} = \sqrt{3} ({\cos 6}^{\circ} + {\sin 12}^{\circ})

\frac{1}{2} + \frac{1}{2} {\sin 6}^{\circ} + \frac{1}{2} {\cos 12}^{\circ} = \frac{\sqrt{3}}{2} {\cos 6}^{\circ} + \frac{\sqrt{3}}{2} {\sin 12}^{\circ}

\frac{1}{2} + (\frac{1}{2} {\sin 6}^{\circ} - \frac{\sqrt{3}}{2} {\cos 6}^{\circ}) + (\frac{1}{2} {\cos 12}^{\circ} - \frac{\sqrt{3}}{2} {\sin 12}^{\circ}) = 0

\frac{1}{2} + \sin (6^{\circ} - 60^{\circ}) + \cos (12^{\circ} + 60^{\circ}) = 0

\frac{1}{2} - {\sin 54}^{\circ} + {\cos 72}^{\circ} = 0

\frac{1}{2} - {\cos 36}^{\circ} + {\sin 18}^{\circ} = 0

\frac{1}{2} - \frac{\sqrt{5} + 1}{4} + \frac{\sqrt{5} - 1}{4} = 0

\frac{1}{2} - \frac{1}{4} - \frac{1}{4} = 0

Leonardo Mcpherson

Beginner2022-04-01Added 13 answers

x = \frac{1 + \sin 6 + \cos 12}{\cos 6 + \sin 12}

= \frac{2 \sin 30 + \sin 78 + \sin 6}{\cos 78 + \cos 6}

= \frac{2 (\sin 54 - \sin 18) + 2 \sin 42 \cos 36}{2 \cos 42 \cos 36}

= \frac{4 \cos 36 \sin 18 + 2 \sin 42 \cos 36}{2 \cos 42 \cos 36}

= \frac{2 \cos 36 (2 \sin 18 + \sin 42)}{2 \cos 42 \cos 36}

= \frac{2 \sin 18 + \sin 42}{\cos 42}

= \frac{\sin 42 + \sin 18 + \sin 18}{\cos 42}

= \frac{2 \sin 30 \cos 12 + \cos 72}{\cos 42}

= \frac{\cos 72 + \cos 12}{\cos 42}

= \frac{2 \cos 42 \cos 30}{\cos 42}

= 2 \cos 30

= \sqrt{3}

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