Answer
$\cot18^{\circ}=\frac{\sqrt {10+2\sqrt 5}}{\sqrt 5-1} ≈ 3.077684$
Work Step by Step
Recall that
$\cos^2x + \sin^2x = 1$
$\cos x = \sqrt {1-\sin^2x}$
To find the value of $\cot 18^{\circ}$ we'll use the next identity:
$\cot x = \frac{\cos x}{\sin x}$
We are given the value of $\sin 18^{\circ}$, so we only have to find the value of $\cos 18^{\circ}$
$\cos 18^{\circ} = \sqrt {1-(\sin 18^{\circ})^2}$
$\cos 18^{\circ} = \sqrt {1-(\frac{\sqrt 5-1}{4})^2}$
$\cos 18^{\circ} = \sqrt {1-(\frac{6-2\sqrt 5}{16})}$
$\cos 18^{\circ} = \sqrt {\frac{10+2\sqrt 5}{16}}$
$\cos 18^{\circ} = {\frac{\sqrt {10+2\sqrt 5}}{4}}$
Find the value of $\cot 18^{\circ}$:
$\cot 18^{\circ}=\frac{\cos 18^{\circ}}{\sin 18^{\circ}}$
$\cot 18^{\circ}=\frac{{\frac{\sqrt {10+2\sqrt 5}}{4}}}{\frac{\sqrt 5-1}{4}}$
$\cot 18^{\circ}= \frac{\sqrt {10+2\sqrt 5}}{\sqrt 5-1}$
$\cot18^{\circ}≈ 3.077684$
