Simplfy trigonometric functions by only considering integer inputs? I have the below function which only takes integer input,
$$ 2 \sqrt{3} \sin \left(\frac{\pi  t}{3}\right)+\sqrt{3} \sin \left(\frac{2 \pi  t}{3}\right)-\sqrt{3} \sin \left(\frac{4 \pi  t}{3}\right)+6 \cos \left(\frac{\pi  t}{3}\right)+\cos \left(\frac{2 \pi  t}{3}\right)+\cos \left(\frac{4 \pi  t}{3}\right) $$
The resulting output from $t=0$ to $t=10$ is $\{8, 8, -4, -4, -4, -4, 8, 8, -4, -4, -4\}$, and this pattern repeats indefinitely.
Is there a systematic way of reducing this expression using the fact we only consider integer input?
 A: Notice that 
$$\begin{align}\sin\left(\frac{4\pi t}{3}\right) &= \sin\left(\frac{6\pi t}{3} - \frac{2\pi t}{3}\right)\\
&= \sin\left(2\pi t - \frac{2\pi t}{3}\right)\\
&= - \sin \left(\frac{2\pi t}{3}\right)\end{align}$$
For the cosine part,
$$\begin{align}\cos\left(\frac{4\pi t}{3}\right) &= \cos\left(\frac{6\pi t}{3} - \frac{2\pi t}{3}\right)\\
&= \cos\left(2\pi t - \frac{2\pi t}{3}\right)\\
&= \cos \left(\frac{2\pi t}{3}\right)\end{align}$$
If we apply these two identitiesto the given expression, then we have
$$ 2 \sqrt{3} \sin \left(\frac{\pi  t}{3}\right)+2\sqrt{3} \sin \left(\frac{2 \pi  t}{3}\right)+6 \cos \left(\frac{\pi  t}{3}\right)+2\cos \left(\frac{2 \pi  t}{3}\right)\\ $$
With the so-called 'R-Formulae', this becomes
$$\sqrt{48}\sin\left(\frac{\pi t}{3} + \arctan{\frac{3}{\sqrt{3}}}\right) + \sqrt{16}\sin\left(\frac{2\pi t}{3} + \arctan{\frac{1}{\sqrt{3}}}\right)\\
= {4\sqrt{3}\sin\left(\frac{\pi t}{3} + \frac{\pi}{3}\right) + 4\sin\left(\frac{2\pi t}{3} + \frac{\pi}{6}\right)}$$
$$= {4\sqrt{3}\sin\left(\frac{(t + 1)\pi t}{3}\right) + 4\sin\left(\frac{(4t + 1)\pi t}{6}\right)}$$
... and this is as far as I got.
I am unsure what you mean by "systematic". If you are referring to a general approach of summing up $\sin$'s and $\cos$'s that does not depend on one's observation skills, then perhaps you can make use of complex numbers.
Sum up the sines by summing up their corresponding complex numbers and taking the imaginary part of the result, and then do the same for the cosines, this time taking the real part of the result.
A: Since $t$ is an integer,
$$ \sin(\tfrac{4\pi t}{3}) = \sin(2\pi t - \tfrac{2\pi t}{3})
= \sin(2\pi t)\cos(\tfrac{2\pi t}{3}) - \cos(2\pi t)\sin(\tfrac{2\pi t}{3})
= -\sin(\tfrac{2\pi t}{3}) $$
and similarly
$$ \cos(\tfrac{4\pi t}{3}) = \cos(\tfrac{2\pi t}{3}) $$
Thus
\begin{align*}
& 2\sqrt3 \sin(\tfrac{\pi t}{3})
+ \sqrt3\sin(\tfrac{2\pi t}{3})
- \sqrt3\sin(\tfrac{4\pi t}{3})
+ 6\cos(\tfrac{\pi t}{3})
+ \cos(\tfrac{2\pi t}{3})
+ \cos(\tfrac{4\pi t}{3}) \\
&= 2\sqrt3 \sin(\tfrac{\pi t}{3}) + 6\cos(\tfrac{\pi t}{3})
+ 2\sqrt3\sin(\tfrac{2\pi t}{3}) + 2\cos(\tfrac{2\pi t}{3}) \\
&= 4\sqrt3 \left(\tfrac12 \sin(\tfrac{\pi t}{3})
  + \tfrac{\sqrt3}{2}\cos(\tfrac{\pi t}{3})\right)
+ 4 \left(\tfrac{\sqrt3}{2}\sin(\tfrac{2\pi t}{3})
  + \tfrac12\cos(\tfrac{2\pi t}{3}) \right) \\
&= 4\sqrt3 \sin(\tfrac{\pi (t+1)}{3}) + 4 \cos(\tfrac{\pi (2t-1)}{3}) \\
&= 4\sqrt3 \sin(\tfrac{\pi (t+1)}{3}) - 4 \cos(\tfrac{\pi (2t-1)}{3}+\pi) \\
&= 4\sqrt3 \sin(\tfrac{\pi (t+1)}{3}) - 4 \cos(\tfrac{2\pi (t+1)}{3}) \\
&= 4\sqrt3 \sin(\tfrac{\pi (t+1)}{3}) - 4 \cos\big(\pi(t+1)-\tfrac{\pi (t+1)}{3}\big)\\
&= 4\sqrt3 \sin(\tfrac{\pi (t+1)}{3}) - (-1)^{t+1} 4 \cos(\tfrac{\pi (t+1)}{3})\\
&= 8\left(\tfrac{\sqrt3}{2} \sin(\tfrac{\pi (t+1)}{3}) + (-1)^t\tfrac12\cos(\tfrac{\pi (t+1)}{3})\right)\\
&= 8\sin(\tfrac\pi6(2t+2+(-1)^t))
\end{align*}
