<s>[INST] How to prove $\cos \frac{2\pi }{5}=\frac{-1+\sqrt{5}}{4}$? I would like to find the apothem of a regular pentagon. It follows from 
$$\cos \dfrac{2\pi }{5}=\dfrac{-1+\sqrt{5}}{4}.$$
But how can this be proved (geometrically or trigonometrically)? 
 [/INST] 
Consider a $\triangle ABC$ with $AB=1$, $\mathrm{m}\angle A=\frac{\pi}{5}$ and $\mathrm{m}\angle B=\mathrm{m}\angle C=\frac{2\pi}{5}$, and point $D$ on $\overline{AC}$ such that $\overline{BD}$ bisects $\angle ABC$.  Now, $\mathrm{m}\angle CBD=\frac{\pi}{5}$ and $\mathrm{m}\angle BDC=\frac{2\pi}{5}$, so $\triangle ABC\sim\triangle BCD$.  Also note that $\triangle ABD$ is isosceles so that $BC=BD=AD$.
Let $x=BC=BD=AD$.  From the similar triangles, $\frac{AB}{BC}=\frac{BC}{CD}$ or $\frac{1}{x}=\frac{x}{1-x}$, so $1-x=x^2$ and $x=\frac{\sqrt{5}-1}{2}$ (the other solution is negative and lengths cannot be negative).
Now, apply the Law of Cosines to $\triangle ABC$:
$$\begin{align}
\cos\frac{2\pi}{5}=\cos C&=\frac{a^2+b^2-c^2}{2ab}
\\\\
&=\frac{\left(\frac{\sqrt{5}-1}{2}\right)^2+1^2-1^2}{2\cdot\frac{\sqrt{5}-1}{2}\cdot 1}
\\\\
&=\frac{\frac{\sqrt{5}-1}{2} \cdot \frac{\sqrt{5}-1}{2}}{2\cdot\frac{\sqrt{5}-1}{2}}
\\\\
&=\frac{\sqrt{5}-1}{4}.
\end{align}$$
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