"Prove that(3−4sin⁡(2π15))cos⁡(π30)=sin⁡(π30)" was answered by our community

Rowan Callahan

Answered question

2022-04-01

Prove that

(\sqrt{3} - 4 \sin (\frac{2 π}{15})) \cos (\frac{π}{30}) = \sin (\frac{π}{30})

Cassius Villarreal

Beginner2022-04-02Added 11 answers

(\sqrt{3} - 4 {\sin 24}^{\circ}) {\cos 6}^{\circ} = {\sin 6}^{\circ}

will be true

\Rightarrow {\tan 60}^{\circ} - {\tan 6}^{\circ} = 4 {\sin 24}^{\circ}

\Rightarrow \sin (60^{\circ} - 6^{\circ}) = 4 {\sin 24}^{\circ} {\cos 60}^{\circ} {\cos 6}^{\circ}

\Rightarrow {\sin 54}^{\circ} = 2 {\sin 24}^{\circ} {\cos 6}^{\circ}

{\cos 60}^{\circ} = \frac{1}{2}

Using Werner Formulas, the Right Hand Side becomes

{\sin 30}^{\circ} + {\sin 18}^{\circ}

So, we need to show

{\sin 54}^{\circ} - {\sin 18}^{\circ} = \frac{1}{2}

{\sin 30}^{\circ} = \frac{1}{2}

Multiplying

S = {\sin 54}^{\circ} - {\sin 18}^{\circ} = {\sin 54}^{\circ} + \sin (- 18^{\circ})

2 \sin (\frac{54^{\circ} - (- 18^{\circ})}{2}) = 2 {\sin 36}^{\circ}

2 {\sin 36}^{\circ} \cdot S = 2 {\sin 36}^{\circ} \cdot {\sin 54}^{\circ} + 2 {\sin 36}^{\circ} \sin (- 18^{\circ})

Applying Werner Formula in the Right Hand Side,

2 {\sin 36}^{\circ} \cdot S = {\cos 18}^{\circ} - {\cos 90}^{\circ} + {\cos 54}^{\circ} - {\cos 18}^{\circ}

\Rightarrow S = \frac{{\cos 54}^{\circ}}{2 {\sin 36}^{\circ}} = \frac{1}{2}

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