Question

$\cos {36}^{\circ }-\cos {72}^{\circ }=$

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

Answer 1

Answer: (B)

$\frac{1}{2}$

$\cos {36}^{\circ }-\cos {72}^{\circ }=-2\sin {54}^{\circ }\sin (-{18}^{\circ })$

$[\cos A-\cos B=2\sin \left(\frac{D-C}{2}\right)\left(\sin \frac{C+D}{2}\right)]$

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

$=\frac{2}{4}=\frac{1}{2}$