How prove this $S_{n}\neq 0$ let $n$ is postive integer numbers,and let
$$S_{n}=\sin{1}-\sin{4}+\cdots+(-1)^{n-1}\sin{(3n-2)}$$
show that
$$S_{n}\neq 0, \forall n\in N^{+}$$
My try: maybe this problem use
$$2\sin{x}\sin{y}=\cos{(x-y)}-\cos{(x+y)}$$
\begin{align*}2\sin{\dfrac{3}{2}}S_{n}&=2\sin{1}\sin{\dfrac{3}{2}}-2\sin{4}\sin{\dfrac{3}{2}}+\cdots+2(-1)^{n-1}\sin{(3n-2)}\sin{\dfrac{3}{2}}\\
&=\cos{\dfrac{1}{2}}-\cos{\dfrac{5}{2}}-\cos{\dfrac{5}{2}}+\cos{\dfrac{11}{2}}+\cdots
\end{align*}
I think  this is nice problem, Thank you


go with product to sum 
then calculate sum of $n$th of sigma
at then end you have $6n-7=+-1+4k\pi$
so
$6k-8=4k\pi$ divide by 4 -->$k$ is not an integer 
$6k-6=4k\pi$ divide by 4 -->$k$ is not an integer 
so $s(n)$ is not zero
