Question

Evaluate : $cos{18}^o$.

Answer 1

$\sin 18=\frac{\sqrt{5}-1}{4}$

$\cos 18=\sqrt{1-{\sin }^{2}18}$

$=\sqrt{1-{\left(\frac{\sqrt{5}-1}{4}\right)}^2}$

$=\sqrt{\left(\frac{16-(5+1-2\sqrt{5})}{16}\right)}$

$=\sqrt{\left(\frac{10+2\sqrt{5}}{16}\right)}$

$\cos 18=\frac{\sqrt{\left(10+2\sqrt{5}\right)}}{4}$