Trig

Recall the trigonometric identities $\begin{aligned} \cos(A) & = & -\cos(\pi - A) \\ 1 - 2\sin^2 (A) = 2 \cos^2 (A) - 1 & = & \cos(2A) \\ 2 \cos(A) \sin(B) & = & \sin \left ( {A+B} \right ) - \sin \left ( {A-B} \right ) \\ 2 \sin(A) \sin(B) & = & -\cos \left ( {A+B} \right ) + \cos \left ( {A-B} \right ) \\ \end{aligned}$

For simplicities sake, let $x = \frac {\pi}{11}$ .

Lemma $\cos(2x) + \cos(4x) + \cos(6x) + \cos(8x) + \cos(10x) = -\frac {1}{2}$

Proof of Lemma

By recurrence relation of Tchebyshev polynomial, $T_{n+1} (y) = 2y \space T_n (y) - T_{n-1} (y)$

So $T_1 (y) = y, T_2 (y) = 2y^2 - 1, T_3 (y)= 4y^3 - 3y, T_4 (y)= 8y^4 - 8y^2 + 1, T_5 (y) = 16y^5 - 20y^4 + 5y , \ldots$

$\cos(11x) = -1$ , state $T_{11} (y) =-1$ in terms of linear combination of $T_1 (y), T_3 (y), T_5 (y), T_7 (y), T_9 (y)$ , we get $T_1 (y) + T_3 (y) + T_5 (y)+ T_7 (y)+ T_9 (y) = \frac {1}{2}$ . With $\cos(A) = - \cos(\pi - A)$ . This completes the proof.

We claim that the answer is $11$ . Proof by contradiction. Suppose otherwise,

$\begin{aligned} 11 \cos^2 (3x) & \ne & \left ( \sin(3x) + 4 \cos(3x) \sin (2x) \right )^2 \\ 11 \cdot \frac { \cos(6x) + 1}{2} & \ne & \left ( \sin(3x) + 2 \sin(5x) - 2\sin(x) \right )^2 \\ 11 ( \cos(6x) + 1) & \ne & 2 \left ( \sin^2 (3x) + 4\sin^2(5x) + 4 \sin^2 (x) + 4 \sin(5x) \sin(3x) - 4\sin(3x) \sin(x) - 8\sin(5x) \sin(x) \right) \\ 11 \cos(6x) + 11 & \ne & 2 \left ( \frac {1 - \cos(6x)}{2} + 2(1 - \cos(10x)) + 2\cos(2x) + 2(-\cos(8x) + \cos(2x) ) - 2(-\cos(4x) + \cos(2x)) - 4(-\cos(6x) + \cos(4x) ) \right) \\ -2 & \ne & 4 \cos(10x) + 4\cos(8x) + 4\cos(6x) + 4\cos(4x) + 4\cos(2x) \\ \end{aligned}$

Which is a contradiction. So our assumption is wrong. Hence the answer is indeed $\boxed{11}$

Nice, simple problem for once! I just rearranged using identities, and arrived at $\boxed{11}$ . Tag me in a cooler solution, whoever writes one. :D