Prove or disprove this : $\forall (k.m)\in \mathbb Z^*\times\mathbb Z^*, \text{and } (z,r,s)\in \mathbb N\times\mathbb N^*\times\mathbb N^*\text{ where } 1\leq r <s$: $$\sin(k\pi^{r+1}+m\pi^{s+1}+z(e\pi+1))\neq0$$
My atempt:
for $z=0$ : $$\sin(k\pi^{r+1}+m\pi^{s+1}+z(e\pi+1))=\sin(k\pi^{r+1}+m\pi^{s+1})\neq0$$
Because $$:k\pi^{r+1}+m\pi^{s+1}\neq a\pi ,\quad a\in \mathbb Z $$
for $z>0$ :
\begin{align*}
\sin(k\pi^{r+1}+m\pi^{s+1}+z(e\pi+1))&=\sin(k\pi^{r+1}+m\pi^{s+1})\cos(ze\pi+z)-\sin(ze\pi+z)\cos(k\pi^{r+1}+m\pi^{s+1})\\
&=\sin(k\pi^{r+1}+m\pi^{s+1})\sqrt{1-\sin^2(ze\pi+z)}-\sin(ze\pi+z)\sqrt{1-\sin^2(k\pi^{r+1}+m\pi^{s+1})}\\
&=\sin(k\pi^{r+1}+m\pi^{s+1})\times\alpha_1-\alpha_2\times\alpha_3\\
\end{align*}
where:$\{_{\alpha_1=\sqrt{1-\alpha_2^2}}^{\alpha_2=\sin(z(e\pi+1))}$
and
$\alpha_3=\sqrt{1-\sin^2(k\pi^{r+1}+m\pi^{s+1})}$
After the calculation, I found that this is true $$\sin(k\pi^{r+1}+m\pi^{s+1})\neq0$$ If this is true $$ \sin(k\pi^{r+1}+m\pi^{s+1})\neq \sqrt{\alpha_2^2(\alpha_3^2+\sin^2(k\pi^{r+1}+m\pi^{s+1})}$$ But now I have no idea to prove this، I posted this question in order to find a suggestion of an idea or a guess about proving this if possible to help me. Thanks in advance .