Buktikan bahwa: 4 sin 20 sin 40 sin 80 = 1/2 akar(3)

Berdasarkan identitas trigonometri, $4 \sin 20^{\circ} \sin 40^{\circ} \sin 80^{\circ}$ terbukti sama dengan $\frac{\sqrt{3}}{2}$.

Pembahasan

Untuk membuktikan $\small 4 \sin 20^{\circ} \sin 40^{\circ} \sin 80^{\circ} = \frac{1}{2} \sqrt{3}$, kita dapat menggunakan identitas trigonometri. Salah satu pendekatan adalah dengan menggunakan rumus perkalian menjadi penjumlahan atau pengurangan, atau dengan menggunakan sifat-sifat fungsi sinus. 

Menggunakan identitas $\small \sin A \sin B = \frac{1}{2} [\cos(A-B) - \cos(A+B)]$: 

$\small \sin 20^{\circ} \sin 40^{\circ} = \frac{1}{2} [\cos(20^{\circ}-40^{\circ}) - \cos(20^{\circ}+40^{\circ})] = \frac{1}{2} [\cos(-20^{\circ}) - \cos(60^{\circ})] = \frac{1}{2} [\cos 20^{\circ} - \frac{1}{2}]$ 

Sekarang, kalikan dengan $\small \sin 80^{\circ}$: 

$\small 4 \sin 20^{\circ} \sin 40^{\circ} \sin 80^{\circ} = 4 \times \frac{1}{2} [\cos 20^{\circ} - \frac{1}{2}] \sin 80^{\circ} = 2 [\cos 20^{\circ} \sin 80^{\circ} - \frac{1}{2} \sin 80^{\circ}]$ 

Menggunakan identitas $\small \cos A \sin B = \frac{1}{2} [\sin(A+B) - \sin(A-B)]$: 

$\small \cos 20^{\circ} \sin 80^{\circ} = \frac{1}{2} [\sin(20^{\circ}+80^{\circ}) - \sin(20^{\circ}-80^{\circ})] = \frac{1}{2} [\sin(100^{\circ}) - \sin(-60^{\circ})] = \frac{1}{2} [\sin 100^{\circ} + \sin 60^{\circ}]$ 

Karena $\small \sin 100^{\circ} = \sin(180^{\circ}-80^{\circ}) = \sin 80^{\circ}$, maka: 

$\small \cos 20^{\circ} \sin 80^{\circ} = \frac{1}{2} [\sin 80^{\circ} + \frac{\sqrt{3}}{2}]$ 

Substitusikan kembali ke persamaan awal: 

$2 [\frac{1}{2} [\sin 80^{\circ} + \frac{\sqrt{3}}{2}] - \frac{1}{2} \sin 80^{\circ}] = 2 [\frac{1}{2} \sin 80^{\circ} + \frac{\sqrt{3}}{4} - \frac{1}{2} \sin 80^{\circ}] = 2 [\frac{\sqrt{3}}{4}] = \frac{\sqrt{3}}{2}$ 

Alternatif lain: Gunakan identitas $\small \sin(3x) = 3\sin x - 4\sin^3 x$. Atau $\small \sin x = \cos(90-x)$. 

$\small \sin 80^{\circ} = \cos(90^{\circ}-80^{\circ}) = \cos 10^{\circ}$. 
$\small \sin 40^{\circ} = \cos(90^{\circ}-40^{\circ}) = \cos 50^{\circ}$. 
$\small \sin 20^{\circ} = \cos(90^{\circ}-20^{\circ}) = \cos 70^{\circ}$. 

$\small 4 \cos 70^{\circ} \cos 50^{\circ} \cos 10^{\circ}$ 

Gunakan identitas $\small \cos A \cos B = \frac{1}{2}[\cos(A-B) + \cos(A+B)]$. 

$\small \cos 70^{\circ} \cos 50^{\circ} = \frac{1}{2}[\cos(70^{\circ}-50^{\circ}) + \cos(70^{\circ}+50^{\circ})] = \frac{1}{2}[\cos 20^{\circ} + \cos 120^{\circ}] = \frac{1}{2}[\cos 20^{\circ} - \frac{1}{2}]$ 

$\small 4 \times \frac{1}{2}[\cos 20^{\circ} - \frac{1}{2}] \cos 10^{\circ} = 2[\cos 20^{\circ} \cos 10^{\circ} - \frac{1}{2} \cos 10^{\circ}]$ 

$\small \cos 20^{\circ} \cos 10^{\circ} = \frac{1}{2}[\cos(20^{\circ}-10^{\circ}) + \cos(20^{\circ}+10^{\circ})] = \frac{1}{2}[\cos 10^{\circ} + \cos 30^{\circ}] = \frac{1}{2}[\cos 10^{\circ} + \frac{\sqrt{3}}{2}]$ 

$2[\frac{1}{2}(\cos 10^{\circ} + \frac{\sqrt{3}}{2}) - \frac{1}{2} \cos 10^{\circ}] = 2[\frac{1}{2} \cos 10^{\circ} + \frac{\sqrt{3}}{4} - \frac{1}{2} \cos 10^{\circ}] = 2[\frac{\sqrt{3}}{4}] = \frac{\sqrt{3}}{2}$.