## Trigonometry (11th Edition) Clone

$tan~18^{\circ} = \frac{\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}} = 0.325$
$cos^2~x+sin^2~x = 1$ $cos~x = \sqrt{1-sin^2~x}$ 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 $tan~18^{\circ}$: $tan~18^{\circ} = \frac{sin~18^{\circ}}{cos~18^{\circ}}$ $tan~18^{\circ} = \frac{\frac{\sqrt{5}-1}{4}}{\frac{\sqrt{10+2\sqrt{5}}}{4}}$ $tan~18^{\circ} = \frac{\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}}$ $tan~18^{\circ} = 0.325$