Trig Power Challenge

We know that:

$\cos{\frac{\pi}{6}} = \frac {\sqrt{3}}{2}\quad \Rightarrow 2\cos^2{\frac{\pi}{12}} - 1 = \frac {\sqrt{3}}{2}$

$\Rightarrow 2(2\cos^2{\frac{\pi}{24}} -1)^2 - 1= \frac {\sqrt{3}}{2}$

$\Rightarrow 8\cos^4{\frac{\pi}{24}}-8\cos^2{\frac{\pi}{24}} + 1 = \frac {\sqrt{3}}{2}$

$\Rightarrow 16\cos^4{\frac{\pi}{24}}-16\cos^2{\frac{\pi}{24}} + 2 - \sqrt{3} = 0$

$\Rightarrow \cos^2{\frac{\pi}{24}} = \frac {2+\sqrt{2+\sqrt{3}}}{4}\quad \Rightarrow \sin^2{\frac{\pi}{24}} = \frac {2-\sqrt{2+\sqrt{3}}}{4}$

Now,

$\sin^{24}{\frac{\pi}{24}} + \cos^{24}{\frac{\pi}{24}} = (\sin^2{\frac{\pi}{24}})^{12} + (\cos^2{\frac{\pi}{24}})^{12}$

$= \left( \frac {2-\sqrt{2+\sqrt{3}}}{4}\right)^{12} + \left( \frac {2+\sqrt{2+\sqrt{3}}}{4}\right)^{12}$

$= \dfrac{1}{4^{12}} \left[ \left( 2-\sqrt{2+\sqrt{3}} \right)^{12} + \left( 2+\sqrt{2+\sqrt{3}} \right)^{12} \right]$

$= \dfrac{2}{4^{12}} \left[ 2^{12} + \begin{pmatrix} 12 \\ 2 \end{pmatrix} (2^{10}) (\sqrt{2+\sqrt{3}})^{2} + \begin{pmatrix} 12 \\ 4 \end{pmatrix} (2^{8}) (\sqrt{2+\sqrt{3}})^{4} \\ \quad +\begin{pmatrix} 12 \\ 6 \end{pmatrix} (2^{6}) (\sqrt{2+\sqrt{3}})^{6} + \begin{pmatrix} 12 \\ 8 \end{pmatrix} (2^{4}) (\sqrt{2+\sqrt{3}})^{8} \\ \quad + \begin{pmatrix} 12 \\ 10 \end{pmatrix} (2^{2}) (\sqrt{2+\sqrt{3}})^{10} + \begin{pmatrix} 12 \\ 12 \end{pmatrix} (\sqrt{2+\sqrt{3}})^{12} \right]$

$= \dfrac{2}{4^{12}} \left[ 2^{12} + \begin{pmatrix} 12 \\ 2 \end{pmatrix} (2^{10}) (2+\sqrt{3}) + \begin{pmatrix} 12 \\ 4 \end{pmatrix} (2^{8}) (2+\sqrt{3})^{2} \\ \quad +\begin{pmatrix} 12 \\ 6 \end{pmatrix} (2^{6}) (2+\sqrt{3})^{3} + \begin{pmatrix} 12 \\ 8 \end{pmatrix} (2^{4}) (2+\sqrt{3})^{4} \\ \quad + \begin{pmatrix} 12 \\ 10 \end{pmatrix} (2^{2}) (2+\sqrt{3})^{5} + \begin{pmatrix} 12 \\ 12 \end{pmatrix} (2+\sqrt{3})^{6} \right]$

$= \dfrac{1}{8388608} \left[ 4096 + (66)(1024)(2+\sqrt{3}) +(495)(256)(7+4\sqrt{3}) \\ \quad +(924)(64)(26+15\sqrt{3}) +(495)(16)(97+56\sqrt{3}) \\ \quad +(66)(4)(362+209\sqrt{3}) + (1351+780\sqrt{3}) \right]$

$= \dfrac {3428999+1960980\sqrt{3}}{8388608}$

$\Rightarrow a+b+c+d = 13778590$ and the last three digits $\boxed{590}$ .

Let k=pi/24.Form a quadratic equation whose roots are (sink)^2 and (cosk)^2.Now apply Newton sum to evaluate sum of twelth power of the roots.The calculations are very long and it requires patience to finally solve the question correctly.