Show that $4\cos\frac\pi5\cdot\sin\frac\pi{10}=1$ 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?

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