Soviet Russian Cosine #1

First note that $\cos(\frac{\pi}{5}) - \cos(\frac{2\pi}{5}) = \cos(\frac{\pi}{5}) + \cos(\frac{3\pi}{5}).$

Now, using the identity $\cos(A) + \cos(B) = 2\cos(\frac{A + B}{2})\cos(\frac{A - B}{2})$ we see that

$\cos(\frac{\pi}{5}) + \cos(\frac{3\pi}{5}) = 2\cos(\frac{2\pi}{5})\cos(\frac{\pi}{5}) =$

$\dfrac{2\cos(\frac{2\pi}{5})\cos(\frac{\pi}{5})\sin(\frac{\pi}{5})}{\sin(\frac{\pi}{5})} = \dfrac{\cos(\frac{2\pi}{5})\sin(\frac{2\pi}{5})}{\sin(\frac{\pi}{5})} = \dfrac{\frac{1}{2}\sin(\frac{4\pi}{5})}{\sin(\frac{\pi}{5})} = \dfrac{1}{2},$

since $\sin(\frac{4\pi}{5}) = \sin(\frac{\pi}{5}).$

Thus our answer can be written as $\dfrac{1*\sqrt{1}}{2},$ and so $a + b + c = 1 + 1 + 2 = \boxed{4}.$

Note that $~~\sin\alpha=\sin\left(\pi-\alpha\right)$

$\begin{array}{c}&2\cos\frac{\pi}{5}~\sin\frac{\pi}{5}-2\cos\frac{2\pi}{5}~\sin\frac{\pi}{5}&=\sin\frac{2\pi}{5}-\left(\sin\frac{3\pi}{5}-\sin\frac{\pi}{5}\right)\\\\2\sin\frac{\pi}{5}\left(\cos\frac{\pi}{5}-\cos\frac{2\pi}{5}\right)&=\sin\frac{2\pi}{5}-\sin\frac{2\pi}{5}+\sin\frac{\pi}{5}=\sin\frac{\pi}{5}\\\\\cos\frac{\pi}{5}-\cos\frac{2\pi}{5}&=\frac{1}{2}=\frac{1\times\sqrt{1}}{2}\end{array}$

Thus, the required answer is $~~a+b+c=1+1+2=\boxed{4}$

Because $\cos(\pi - A) = - \cos(A)$ , $\cos \left ( \frac {\pi}{5} \right ) - \cos \left ( \frac {2\pi}{5} \right ) = \cos \left ( \frac {\pi}{5} \right ) + \cos \left ( \frac {3\pi}{5} \right )$

I claim the answer is $\boxed{\frac 1 2}$ .

Proof : Let $\omega = \cos \left( \frac {2\pi}{5} \right)+ i \sin \left( \frac {2\pi}{5}\right)$ , then $1+\omega +\omega^2+\omega^3+\omega^4 = 0$ . Comparing real parts

$1 + \cos \left(\frac {2\pi}{5}\right)+ \cos \left(\frac {4\pi}{5} \right)+ \cos \left(\frac {6\pi}{5} \right)+ \cos\left( \frac {8\pi}{5} \right)= 0 \\ \Rightarrow 2\cos \left( \frac {\pi}{5} \right)+ 2 \cos \left( \frac {3\pi}{5}\right) = 1$