How prove this $S_{n}\neq 0$

Question

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

Answer 1

From Euler's formula $e^{i\theta}=\cos\theta+i\sin\theta$, we know $\sin\theta=\Im~ e^{i\theta}$. Then $$S_n=\sum_{k=1}^n(-1)^{k-1}\sin(3k-2)=\Im~\sum_{k=1}^n(-1)^{k-1}e^{i(3k-2)}$$ The right hand side is actually the sum of geometric series with common ratio $-e^{3i}$. We have $$S_n=\Im\frac{e^i(1-(-e^{3i})^n)}{1-(-e^{3i})}=\Im\frac{e^i-(-1)^ne^{i(3n+1)}}{1+e^{3i}}\neq0$$ The last step is a direct consequence of rationalize denominator of complex number.

Answer 2

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