Sines and Sums and Squares

This problem is a mess; we have such a complicated equation to solve with $2014$ variables, we have to find all solutions, then find the last three digits. Where should we start?

Well, let's first break the equation up into little parts and look at one of those parts:

$\sin^2(a_1+a_2)=(\sin a_1+\sin a_2)(\cos a_1+\cos a_2)$

Let's expand the left side:

$(\sin a_1 \cos a_2 + \sin a_2 \cos a_1)^2=(\sin a_1+\sin a_2)(\cos a_1+\cos a_2)$

Hey, wait a sec. This looks awfully familiar to the Cauchy Schwarz inequality. Let's apply Cauchy Schwarz on the LHS:

$(\sin a_1 \cos a_2 + \sin a_2 \cos a_1)^2\le (\sin^2 a_1+\sin^2 a_2)(\cos^2 a_1+\cos^2 a_2)$

Thankfully, since $\sin^2 \alpha \le \sin \alpha$ and $\cos^2\alpha \le \cos \alpha$ if $0\le \alpha \le \dfrac{\pi}{2}$ , we have

$(\sin^2 a_1+\sin^2 a_2)(\cos^2 a_1+\cos^2 a_2)\le (\sin a_1+\sin a_2)(\cos a_1+\cos a_2)$

Thus,

$\sin^2(a_1+a_2)=(\sin a_1 \cos a_2 + \sin a_2 \cos a_1)^2\le (\sin a_1+\sin a_2)(\cos a_1+\cos a_2)$

But what are the equality cases? In the Cauchy Schwarz inequality, the equality case is when $\dfrac{\sin a_1}{\cos a_2}=\dfrac{\sin a_2}{\cos a_1}$ . This simplifies to $\sin 2a_1=\sin 2a_2$ which means either $a_1=a_2$ or $a_1,a_2=0\text{ or }\dfrac{\pi}{2}$ .

The equality case for $\sin^2 a_1\le \sin a_1$ is $a_1=0\text{ or }\dfrac{\pi}{2}$ (and ditto for $a_2$ and the $\cos$ functions). Thus, the equality case of the inequality overall is only $a_1,a_2=0\text{ or }\dfrac{\pi}{2}$ .

But now we've basically solved the question: $a_i=0\text{ or }\dfrac{\pi}{2}$ for all $i=1\to 2014$ ; thus, there are $2^{2014}$ different ordered $2014$ -tuples of solutions. All we need to do is find the last three digits of $2^{2014}$ .

This is not hard with Euler's theorem: since $\phi (1000)=400$ , we have that $2^{2014}\equiv 2^{5\cdot 400+14}\equiv 2^{14}\equiv 16384\equiv \boxed{384}\pmod{1000}$ and we are done.