Answer
$tan~18^{\circ} = \frac{\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}} = 0.325$
Work Step by Step
$cos^2~x+sin^2~x = 1$
$cos~x = \sqrt{1-sin^2~x}$
We can find the value of $cos~18^{\circ}$:
$cos~18^{\circ} = \sqrt{1-(sin~18^{\circ})^2}$
$cos~18^{\circ} = \sqrt{1-(\frac{\sqrt{5}-1}{4})^2}$
$cos~18^{\circ} = \sqrt{1-(\frac{6-2\sqrt{5}}{16})}$
$cos~18^{\circ} = \sqrt{\frac{10+2\sqrt{5}}{16}}$
$cos~18^{\circ} = \frac{\sqrt{10+2\sqrt{5}}}{4}$
We can find the value of $tan~18^{\circ}$:
$tan~18^{\circ} = \frac{sin~18^{\circ}}{cos~18^{\circ}}$
$tan~18^{\circ} = \frac{\frac{\sqrt{5}-1}{4}}{\frac{\sqrt{10+2\sqrt{5}}}{4}}$
$tan~18^{\circ} = \frac{\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}}$
$tan~18^{\circ} = 0.325$
