Tentukan nilai bentuk trigonometri berikut.

$\sqrt{3} - 2$

Pembahasan

Untuk menentukan nilai dari $\frac{\cos(75^{\circ})}{\cos(165^{\circ})}$, kita dapat menggunakan identitas trigonometri. Pertama, kita perlu mengetahui nilai $\cos(75^{\circ})$ dan $\cos(165^{\circ})$.

Kita tahu bahwa $\cos(75^{\circ}) = \cos(45^{\circ} + 30^{\circ}) = \cos(45^{\circ})\cos(30^{\circ}) - \sin(45^{\circ})\sin(30^{\circ})$.
$\, = \left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) - \left(\frac{\sqrt{2}}{2}\right)\left(\frac{1}{2}\right) = \frac{\sqrt{6} - \sqrt{2}}{4}$.

Selanjutnya, kita tahu bahwa $\cos(165^{\circ}) = \cos(180^{\circ} - 15^{\circ}) = -\cos(15^{\circ})$.
Dan $\cos(15^{\circ}) = \cos(45^{\circ} - 30^{\circ}) = \cos(45^{\circ})\cos(30^{\circ}) + \sin(45^{\circ})\sin(30^{\circ})$.
$\, = \left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(\frac{\sqrt{2}}{2}\right)\left(\frac{1}{2}\right) = \frac{\sqrt{6} + \sqrt{2}}{4}$.
Jadi, $\cos(165^{\circ}) = -\frac{\sqrt{6} + \sqrt{2}}{4}$.

Sekarang kita hitung perbandingannya:
$\frac{\cos(75^{\circ})}{\cos(165^{\circ})} = \frac{\frac{\sqrt{6} - \sqrt{2}}{4}}{-\frac{\sqrt{6} + \sqrt{2}}{4}} = -\frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} + \sqrt{2}}$.
Untuk menyederhanakannya, kita kalikan dengan sekawan penyebut:
$= -\frac{(\sqrt{6} - \sqrt{2})(\sqrt{6} - \sqrt{2})}{(\sqrt{6} + \sqrt{2})(\sqrt{6} - \sqrt{2})} = -\frac{(\sqrt{6})^2 - 2\sqrt{6}\sqrt{2} + (\sqrt{2})^2}{(\sqrt{6})^2 - (\sqrt{2})^2}
$= -\frac{6 - 2\sqrt{12} + 2}{6 - 2} = -\frac{8 - 2(2\sqrt{3})}{4} = -\frac{8 - 4\sqrt{3}}{4}
$= -(2 - \sqrt{3}) = \sqrt{3} - 2$.