Question: We have to prove this identity : $$4\cos\frac\pi5\cdot\sin\frac\pi{10}=1$$
The book's hint is we somehow find out that $\displaystyle1+2\left(\cos\frac\pi5-\cos\frac{2\pi}5\right)$ equals something (from the geometry of a regular pentagon) and the result follows. But I can neither see why the book's hint is true nor how it could yield a result for the original problem. So could you help me proving it either by the book's hint or another way?

Answer: 
Given $\displaystyle 2\cos \frac{\pi}{5}\cdot \sin \frac{\pi}{10}\;,$ Now Let $\displaystyle \frac{\pi}{10} = \theta \;,$ and using $$ 2\sin A\cdot \cos A = \sin 2A$$

Then $$\displaystyle 2\cos 2\theta \cdot \sin \theta=\frac{2\cos 2\theta\cdot \sin \theta \cdot \cos \theta}{\cos \theta} = \frac{2\sin 2\theta \cdot \cos 2\theta}{2\cos \theta} =\frac{\sin 4\theta}{2\cos \theta}$$

So we get $$\displaystyle \frac{\sin 4 \theta}{2\cos \theta} = \frac{\sin \left(\frac{4\pi}{10}\right)}{2\cos \frac{\pi}{10}} = \frac{\cos \frac{\pi}{10}}{2\cos \frac{\pi}{10}} = \frac{1}{2}$$

Bcz $$\displaystyle \sin \frac{4\pi}{10} = \sin \left(\frac{\pi}{2}-\frac{\pi}{10}\right) = \cos \frac{\pi}{10}$$
HINT:

Notice $$LHS=2\cos \frac{\pi}{5}\sin\frac{\pi}{10}$$
$$=2\cos 36^\circ \sin18^\circ$$ Setting the values of $\cos 36^\circ$ & $\sin 18^\circ$ 

$$=2\left(\frac{\sqrt 5+1}{4}\right)\left(\frac{\sqrt 5-1}{4}\right)$$
$$=\frac{(\sqrt 5+1)(\sqrt 5-1)}{8}$$
$$=\frac{(\sqrt 5)^2-1^2}{8}$$ $$=\frac{5-1}{8}=\frac{1}{2}$$