Question

$\cos 36^{\circ}-\cos 72^{\circ}$ is equal to

• A. 1
• B. $\frac{1}{2}$
• C. $\frac{1}{4}$
• D. $\frac{1}{8}$

Answer 1

Answer: (B)

$\cos 36^{\circ}-\cos 72^{\circ}=-2 \sin \left(\frac{36^{\circ}+72^{\circ}}{2}\right)$
$\sin \left(\frac{72^{\circ}-36^{\circ}}{2}\right)$
$=-2 \sin 54^{\circ} \sin \left(-18^{\circ}\right)$
$=2 \sin 54^{\circ} \sin 18^{\circ}$
$=2 \times \frac{\sqrt{5}+1}{4} \times \frac{\sqrt{5}-1}{4}=2 \times \frac{5-1}{4 \times 4}=\frac{4}{8}$
$=\frac{1}{2}$