Prove that 

i)  $\mathrm{Sin}{18}^{\circ }=\frac{\sqrt{5-1}}{1}$

ii) $\cos {36}^{\circ }=\frac{\sqrt{5}+1}{4}$

i)  $\mathrm{Sin}{18}^0=\frac{\sqrt{5}-1}{4}$

Let   $A={18}^0\Rightarrow 5A=5\times 18={90}^{\circ }$

$\Rightarrow 2A+3A={90}^{\circ }\Rightarrow 2A={90}^{\circ }-3A$
Apply $‘Sin’$ on both sides 

$\Rightarrow \mathrm{Sin}2A=\mathrm{Sin}\left({90}^{\circ }-3\mathrm{A}\right)$

$\Rightarrow \mathrm{Si}n 2A=Cos 3A$

$\Rightarrow 2\mathrm{Sin} A\mathrm{Cos} A=4{\mathrm{Cos}}^3 A-3\mathrm{Cos} A$

$\Rightarrow 2\mathrm{Sin}A\mathrm{Cos}A=\mathrm{Cos}A\left(4{\mathrm{Cos}}^2 A-3\right)$

$\Rightarrow 2\mathrm{Si}n A=4 {\mathrm{Cos}}^{2}A-3$

$\Rightarrow 2\mathrm{Sin}A=4\left(1-{\sin }^{2}A\right)-3$

$\Rightarrow 2\mathrm{Si}nA=4-4{\sin }^{2}A-3$

$\Rightarrow 4{\mathrm{Sin}}^{2}A+2\mathrm{Sin}A-1=0$

This is quadratic equation in $Sin A.$

$a=4$ ;   $b=2$;  $c=-1$

$\mathrm{Sin}A=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-2\pm \sqrt{2^2-4.4(-1)}}{2.4}$

$=\frac{-2\pm \sqrt{4+16}}{8}=\frac{-2\pm \sqrt{20}}{8}=\frac{-2\pm 2\sqrt{5}}{8}$

$=\frac{-1\pm \sqrt{5}}{4}$

$\mathrm{Sin}{18}^{\circ }=\frac{-1\pm \sqrt{5}}{4}$  But $\mathrm{Sin}{18}^0>0$

$\therefore$$\mathrm{Sin}{18}^{\circ }=\frac{\sqrt{5}-1}{4}$

ii)  $\mathrm{Cos}{36}^{\circ }=\mathrm{Cos}\left(2\times {18}^{\circ }\right)$

$=1-2{\sin }^{2}18$     $\left(\cos 2\mathrm{A}=1-2{\mathrm{Sin}}^2\mathrm{A}\right)$

$=1-2{\left(\frac{\sqrt{5}-1}{4}\right)}^2=1-2\cdot \frac{(\sqrt{5}-1{)}^2}{16}$

$=\frac{8-(5+1-2\sqrt{5})}{8}=\frac{8-6+2\sqrt{5}}{8}=\frac{2+2\sqrt{5}}{8}$

$=\frac{2(\sqrt{5}+1)}{8}$

$\therefore$$\cos {36}^{\circ }=\frac{\sqrt{5}+1}{4}$