Sum of Fourth Powers of cosine series has closed form solution. A problem posed in the 1988 Irish Mathematical Olympiad asks for a proof of the following
$$\sum\limits_{k=1}^{n} \cos^{4}\Big(\frac{k\pi}{2n+1}\big) = \frac{6n-5}{16}$$ 
Can anyone give me a heads-up on how to proceed to prove this very interesting result ?

Using the hints given in answers, we can make the result more general
$$\color{blue}{\sum\limits_{k=1}^{n} \cos^{2p}\Big(\frac{k\pi}{2n+1}\big) =  4^{-p} \binom{2 p}{p}\,n+4^{-p} \binom{2 p-1}{p}-\frac{1}{2}}$$ and get the beautiful
$$\left(
\begin{array}{cc}
p & \sum\limits_{k=1}^{n} \cos^{2p}\Big(\frac{k\pi}{2n+1}\big) \\
 1 & \frac{2 n-1}{4} \\
 2 & \frac{6 n-5}{16}  \\
 3 & \frac{10 n-11}{32}  \\
 4 & \frac{70 n-93}{256}  \\
 5 & \frac{126 n-193}{512}  \\
 6 & \frac{462 n-793}{2048} \\
 7 & \frac{858 n-1619}{4096} \\
 8 & \frac{12870 n-26333}{65536}
\end{array}
\right)$$
