Finding the $n^{\text{th}}$ term of $-1,-1,-1,-1,1,1,1,1,...$ as a repeating 8-block In my work I came across that sequence
$-1,-1,-1,-1,1,1,1,1,\dots$ and repeating this 8-block so on forever
Now I cant find an ( e.g. trigonometric/complex ) expression
$f(n)$ ( e.g. $f(n) =(-1)^g(n)$ ) which gives me the sequence starting with 
$n=2,3,4,5,6,7,8,9,…$
and so on forever.

By symmetry, we can assume
$$
f(n) =
a_1 \sin\frac{\pi n}{4}
+a_3 \sin\frac{3\pi n}{4}
+b_1 \cos\frac{\pi n}{4}
+b_3 \cos\frac{3\pi n}{4}.
$$
Then
$$
\begin{aligned}
f(1) &= \frac{a_1}{\sqrt{2}} +\frac{a_3}{\sqrt{2}} + \frac{b_1}{\sqrt{2}} -\frac{b_3}{\sqrt{2}} = -1\\
f(2) &= a_1 - a_3 = -1\\
f(3) &= \frac{a_1}{\sqrt{2}} +\frac{a_3}{\sqrt{2}} - \frac{b_1}{\sqrt{2}} +\frac{b_3}{\sqrt{2}} = -1 \\
f(4) &= -b_1 - b_3 = -1.
\end{aligned}
$$
So
$$
\begin{aligned}
a_1 &= -\frac{1}{2} -\frac{1}{\sqrt 2}, \\
a_3 &= +\frac{1}{2} -\frac{1}{\sqrt 2}, \\
b_1 &= b_3 = \frac{1}{2}.
\end{aligned}
$$
So
$$\begin{aligned}
f(n)
&=
\left( -\frac{1}{2} -\frac{1}{\sqrt 2} \right) \sin\frac{\pi n}{4}
+
\left(  \frac{1}{2} -\frac{1}{\sqrt 2} \right) \sin\frac{3 \pi n}{4}
+
\frac{1}{2} \left( \cos\frac{\pi n}{4} + \cos\frac{3 \pi n}{4} \right) \\
&=
\sin\frac{\pi n}{4} \cos\frac{\pi n}{2}
-\sqrt 2 \cos\frac{\pi n}{4} \sin\frac{\pi n}{2}
+
\cos\frac{\pi n}{4} \cos\frac{\pi n}{2} \\
&=
\sqrt 2 \left[ \sin\frac{\pi (n+1)}{4} \cos\frac{\pi n}{2}
- \cos\frac{\pi n}{4} \sin\frac{\pi n}{2} \right].
\end{aligned}
$$
