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)

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