Jack Bauer is back!

By double angle formula, $\frac {1}{2} = \cos \left ( \frac {\pi}{6} \right ) = 2 \cos^2 \left ( \frac {\pi}{12} \right ) - 1 \Rightarrow \cos \left ( \frac {\pi}{12} \right ) = \frac {\sqrt3 + 1}{2\sqrt2} \Rightarrow \sin \left ( \frac {\pi}{12} \right ) = \frac {\sqrt3 - 1}{2\sqrt2}$

$\large \sin^{8} \left ( \frac {\pi}{24} \right ) + \cos^{8} \left ( \frac {\pi}{24} \right )$

$\large = \left ( \sin^2 \left ( \frac {\pi}{24} \right ) \right )^{4} + \left ( \cos^2 \left ( \frac {\pi}{24} \right ) \right )^{4}$

$\large = \left ( \frac {1 - \cos \left (\frac {\pi}{12} \right) }{2} \right )^{4} + \left ( \frac {1 + \cos \left (\frac {\pi}{12} \right) }{2} \right )^{4}$

$\large = \left ( \frac {1 - \frac {1+\sqrt3}{2\sqrt2} }{2} \right )^{4} + \left ( \frac {1 + \frac {1+\sqrt3}{2\sqrt2} }{2} \right )^{4}$

$\large = \frac {1}{2^{10}} \cdot \left ( (2\sqrt2 - (1+\sqrt3))^{4} + ( (2\sqrt2 + (1+\sqrt3))^{12} \right )$

$\large = \frac {1}{2^9} \cdot \left ( (2\sqrt2)^4 + 6 (2\sqrt2)^2 (1+\sqrt3))^{2} + (1+\sqrt3))^{4} \right )$

$\large = \frac {71 + 28 \sqrt3 }{2^7}$

$\begin{aligned} \sin^{24} \left ( \frac {\pi}{24} \right ) + \cos^{24} \left ( \frac {\pi}{24} \right ) & = & \left ( \sin^8 \left ( \frac {\pi}{24} \right ) + \cos^8 \left ( \frac {\pi}{24} \right ) \right )^3 - 3 \cdot \left ( \sin \left ( \frac {\pi}{24} \right ) \cos \left ( \frac {\pi}{24} \right ) \right )^{16} \left ( \sin^8 \left ( \frac {\pi}{24} \right ) + \cos^8 \left ( \frac {\pi}{24} \right ) \right ) \\ & = & \left ( \frac {71 + 28\sqrt3}{2^7} \right )^3 - \frac {3}{2^{8}} \left ( 2 \sin \left ( \frac {\pi}{24} \right ) \cos \left ( \frac {\pi}{24} \right ) \right )^{8} \left ( \frac {71 + 28\sqrt3}{2^7} \right ) \\ & = & \left ( \frac {71 + 28\sqrt3}{2^7} \right )^3 - \frac {3}{2^{8}} \left ( \sin \left ( \frac {\pi}{12} \right ) \right )^{8} \left ( \frac {71 + 28\sqrt3}{2^7} \right ) \\ & = & \left ( \frac {71 + 28\sqrt3}{2^7} \right )^3 - \frac {3}{2^{8}} \left ( \frac {\sqrt3 - 1}{2\sqrt2} \right )^{8} \left ( \frac {71 + 28\sqrt3}{2^7} \right ) \\ & = & \frac { 2^{6} (71 + 28\sqrt3)^3 - 3(71 + 28\sqrt3) (\sqrt3 - 1)^{8} }{2^{27}} \\ \end{aligned}$

We want to expand $( \sqrt3 - 1)^{8}$ , note that $8 = 2^3$ . Consider the binomial expansion $(x + y\sqrt3)^2 = (x^2 + 3y^2) + \sqrt3 (2xy)$

Recurrence relation $(x_{n+1}, y_{n+1} ) = (x_n^2 + 3y_n^2, 2x_n y_n)$ with $(x_0, y_0) = (-1,1)$ , then

$(x_1, y_1) = (4, -2), (x_2, y_2) = (28, -16), (x_3, y_3) = (1552, -896)$ .

Note that $\text{gcd} (x_4,y_4) = 2^3$

$\begin{aligned} \sin^{24} \left ( \frac {\pi}{24} \right ) + \cos^{24} \left ( \frac {\pi}{24} \right ) & = & \frac { 2^{6} (71 + 28\sqrt3)^3 - 3(71 + 28\sqrt3) \cdot 2^4 ( 97 - 56\sqrt3) }{2^{27}} \\ & = & \frac { 2^{2} (71 + 28\sqrt3)^3 - 3(71 + 28\sqrt3) ( 97- 56\sqrt3) }{2^{23}} \\ \end{aligned}$

Expansion of $(71 + 28\sqrt3) ( 97- 56\sqrt3)$ shows that it doesn't have a factor of $2$ . So the numerator is no longer divisible by $2$ . Thus $c$ is minimized. Hence $c = 23 \Rightarrow c+1= \boxed{24}$

Whoa!!!You all forgot EULER's great theorem if pi/24 = t then given expression becomes ( (e^it - e^-it)/2i)^24 + ( (e^it + e^-it)/2)^24

= (2^-24) ( (e^it - e^-it)^24 + (e^it + e^-it)^24 ) since i^24=1

using binomial expansion u will get this to be

{sum 0 to 12 of 24 C 2n e^i(24-2n)t} /2^23

sigma stuff is irrelavent to our Q which requires c(23) hence ANS is 24