<s>[INST] calculate a trigonometric expression related to $\sin(\pi/5)$ I tried to solve a question which led to 
$$4 \sin\frac{\pi}{5} \sin\frac{3\pi}{5} = \sqrt{5}$$
But, I got stuck to prove that.
Is there any easy solution?
 [/INST] I will find the $\cos\frac{2\pi}{5}$ with basic complex number theory. 
Let $z = \cos(\frac{2\pi}{5}) + i \sin(\frac{2\pi}{5})$, then $z^5=1$. (de Moivre's Theorem) Let's find the solution of this quintic equation.
This equation is factored into $(z-1)(z^4 + z^3 + z^2 + z +1 ) = 0$. $z=1$ is trivial solution. 
Now let $u = z + z^4$, and $v= z^2 + z^3$. then $u + v = -1$, and $uv = z^3 + z^4 + z^6 + z^7 = z + z^2 + z^3 + z^4 = -1$. So $t^2 + t - 1 = 0$ has two solutions $u$ and $v$. 
Solving the quadratic equation, we get $t = \frac{-1 \pm \sqrt{5}}{2}$. 
Geometrically $u$ has positive real part, and $v$ has negative real part. so $z + z^4 = \frac{-1 + \sqrt{5}}{2}$. Since $z$ and $z^4$ are conjugate, the real part of $z$ is the half of it, i.e. $\cos\frac{2\pi}{5}=\frac{-1 + \sqrt{5}}{4}$. 
Now, $\cos(\frac{2\pi}{5}) = 1-2\sin^2(\frac{\pi}{5})$, so you can find $\sin\frac{\pi}{5}$.
In this way you can find $\cos\frac{2\pi}{17}$, by decreasing the order of equation 16th to 8th, 4th, .... You need to take $u$ and $v$ in special way...
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