Nice identity

$\begin{aligned} S & = \sum_{n=0}^9 \cos^4 \left(\frac {n\pi}9 \right) \\ & = \sum_{n=0}^9 \left(1-\sin^2 \left(\frac {n\pi}9 \right) \right)^2 \\ & = \sum_{n=0}^9 \left(1- \frac 12 \left(1-\cos \left(\frac {2n\pi}9 \right) \right) \right)^2 \\ & = \frac 14 \sum_{n=0}^9 \left(1 + 2 \cos \left(\frac {2n\pi}9 \right) + \cos^2 \left(\frac {2n\pi}9 \right) \right) \\ & = \frac 14 \sum_{n=0}^9 \left(\frac 32 + 2 \cos \left(\frac {2n\pi}9 \right) + \frac 12 \cos \left(\frac {4n\pi}9 \right) \right) \\ & = \frac 38 \sum_{n=0}^9 1 + \frac 12 \sum_{n=0}^9\cos \left(\frac {2n\pi}9 \right) + \frac 18 \sum_{n=0}^9 \cos \left(\frac {4n\pi}9 \right) & \small \color{#3D99F6} \text{Note that }\sum_{n=0}^9 \cos \left(\frac {4n\pi}9 \right) = \sum_{n=0}^9 \cos \left(\frac {2n\pi}9 \right) \\ & = \frac 38(10) + \frac 58 \sum_{n=0}^9\cos \left(\frac {2n\pi}9 \right) \\ & = \frac {15}4 + \frac 58 \left(\cos 0 +{\color{#3D99F6} \sum_{n=1}^4\cos \left(\frac {2n\pi}9 \right)} + {\color{#D61F06} \sum_{n=5}^8\cos \left(\frac {2n\pi}9 \right)} + \cos 2\pi \right) & \small \color{#3D99F6} \text{See proof: }\sum_{k=1}^n\cos \left(\frac {2n\pi}{2n+1} \right) = - \frac 12 \\ & = \frac {15}4 + \frac 58 \left(1 {\color{#3D99F6} - \frac 12} {\color{#D61F06} - \sum_{n=5}^8\cos \left(\frac {(9-2n)\pi}9 \right)} + 1 \right) & \small \color{#D61F06} \text{Note that }\cos (\pi - \theta) = - \cos \theta \\ & = \frac {15}4 + \frac 58 \left(1 - \frac 12 {\color{#D61F06} - \sum_{n=1}^4\cos \left(\frac {(2n-1)\pi}9 \right)} + 1 \right) & \small \color{#D61F06} \text{See proof: }\sum_{k=1}^n\cos \left(\frac {(2n-1)\pi}{2n+1} \right) = \frac 12 \\ & = \frac {15}4 + \frac 58 \left(1 - \frac 12 {\color{#D61F06} - \frac 12} + 1 \right) \\ & = \frac {15}4 + \frac 58 = \frac {35}8 \end{aligned}$

Therefore, $a+b = 35 + 8 = \boxed{43}$ .

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Proof for $\color{#3D99F6} \displaystyle \sum_{k=1}^n \cos \frac {2k-1}{2n+1}\pi = \frac 12$

My approach

By Euler's Formula we have $e^{ix } = \cos x + i\sin x \implies e^{4 ix } =(\cos x +i\sin x )^4 \\ \Re(e^{4i x} ) = \cos 4x = \cos^4 x -6\cos^2 x \sin^2x +\sin^4 x$ and simplifying the latter expression we have $\cos^4x =\dfrac{\cos 4x +4\cos 2x +3}{8} \\ \sum \cos^4x =\sum\,\dfrac{\cos4x +4\cos 2x +3}{8}\\ \sum_{n=0}^9 \cos^4\left(\frac{n\pi}{9}\right) =\dfrac{1}{8}\sum_{n=0}^9\left(\cos \left(\frac{ 4 n\pi}{9}\right)+3 +4\cos\left(\frac{2n\pi}{9}\right)\right) \\ =\dfrac{1}{8}\sum_{n=0}^{9}\left(\Re(e^{\frac{4ni\pi}{9}})+3 +4\Re (e^{\frac{2ni\pi}{9}}) \right)\\=\dfrac{30+\dfrac{\left(e^{\frac{4ni\pi}{9}}\right)^9 -1}{e^{\frac{4ni\pi}{9}}-1} +4\dfrac{\left(e^{\frac{2ni\pi}{9}}\right)^9 -1}{e^{\frac{2ni\pi}{9}}-1}}{8}\\ =\dfrac{30+1+4}{8}=\dfrac{35}{8}$ and hence $a+b=35+8=43$