Answer
$sin~72^{\circ} = \sqrt{\frac{5+\sqrt{5}}{8}}$
Work Step by Step
$sin~72^{\circ} = sin~(90^{\circ}-18^{\circ}) = cos~18^{\circ}$
$cos^2~x+sin^2~x = 1$
$cos~x = \sqrt{1-sin^2~x}$
We can find the value of $cos~18^{\circ}$:
$cos~18^{\circ} = \sqrt{1-(sin~18^{\circ})^2}$
$cos~18^{\circ} = \sqrt{1-(\frac{\sqrt{5}-1}{4})^2}$
$cos~18^{\circ} = \sqrt{1-(\frac{6-2\sqrt{5}}{16})}$
$cos~18^{\circ} = \sqrt{\frac{10+2\sqrt{5}}{16}}$
$cos~18^{\circ} = \sqrt{\frac{5+\sqrt{5}}{8}}$
Since $sin~72^{\circ} = cos~18^{\circ}$, then $sin~72^{\circ} = \sqrt{\frac{5+\sqrt{5}}{8}}$
