When $m$ is odd, since $\sin (k(\pi-x)) = (-1)^{k-1} \sin (\pi x)$, we only need to verify that the inequalities
$$
\sum_{k=1}^{n} (-1)^{k} \Big( \frac{\sin (kx)}{k} \Big)^m < 0 < \sum_{k=1}^{n} \Big(\frac{\sin (kx)}{k} \Big)^{m}
$$
hold for $0< x\le \pi/2$. If $m$ is even then $(\sin (k(\pi -x)))^2 = (\sin (kx))^2$, and once again it is enough
to verify these inequalities for $0 < x\le \pi/2$.

We start by proving the inequality
$$
F_m(x;n): = \sum_{k=1}^{n} \Big(\frac{\sin (kx)}{k} \Big)^{m} >0.
$$
We may assume that $m\ge 3$, the cases for $m=1$ and $2$ already being known. Suppose that $K\ge 2$ is a natural number
such that $\pi/(K+1) < x \le \pi/K$. Since $\sin(kx) \ge 0$ for all $1\le k\le K$ and $(\sin(kx))^m \ge -1$ for $k \ge K+1$, we find that
$F_m(x;n) \ge 0$ if $n \le K$, and if $n \ge K+1$ then
\begin{align*}
F_m(x;n) &\ge \sum_{k=1}^{K} \Big( \frac{\sin(kx)}{k} \Big)^m - \sum_{k =K+1}^{\infty} \frac{1}{k^m} \\
&\ge \sum_{k=1}^{K} \Big( \frac{\sin(kx)}{k}\Big)^m - \int_{K}^{\infty}
\frac{dt}{t^m} \\
&= \sum_{k=1}^{K} \Big( \frac{\sin(kx)}{k}\Big)^m -\frac{1}{(m-1)K^{m-1}}.
\end{align*}
Note that if $0\le t\le \pi/2$ then $\sin t \ge 2t/\pi$. Using this for $k\le K/2$ above, we conclude that
$$
F_m(x;n) \ge \sum_{k\le K/2} \Big( \frac{2k x}{\pi k} \Big)^m - \frac{1}{(m-1)K^{m-1}} \ge \Big\lfloor \frac K2 \Big\rfloor \Big(\frac{2}{(K+1)}\Big)^m - \frac{1}{(m-1) K^{m-1}}.
$$
Since $K\ge 2$ we have $2/(K+1) \ge 4/(3K)$ and $\lfloor \frac K2 \rfloor \ge K/3$, so that
$$
F_m(x;n) \ge \frac{1}{K^{m-1}} \Big( \frac{1}{3} \Big(\frac 43\Big)^m - \frac{1}{m-1} \Big) > 0
$$
since $m\ge 3$. This completes the proof of one inequality.

We now turn to the other inequality, which is more involved. We rewrite this inequality as
$$
G_m(x;n) := \sum_{k=1}^{n} (-1)^{k-1} \Big( \frac{\sin (kx)}{k} \Big)^m > 0,
$$
which we want to establish for all $m\ge 2$ and all $n$ (the case $m=1$ following from the Jackson inequality).

Once again suppose that $K\ge 2$ is such that $\pi/(K+1) < x\le \pi/K$. Since $\frac{\sin y}{y}$ is decreasing and non-negative in $0 \le y\le \pi$, by pairing
up two consecutive terms we see that $G_m(x;n)$ is positive if $n\le K$ (if there is an unpaired term at the end it is non-negative), and henceforth we assume that $n \ge K+1$. We now record a useful estimate
for tails of our sum, which we shall prove later: for any $0 < x\le \pi/2$ and any integers $A <B$ we have
$$
\Big| \sum_{k=A}^{B} (-1)^{k-1} \Big(\frac{\sin (kx)}{k}\Big)^m \Big| \le \frac{m}{2(m-1)} \frac{x}{A^{m-1}} + \frac{1}{A^m}.
$$
With this estimate in place, we now finish the proof of the inequality. There are two cases $K \ge m$ and $2 \le K <m$ (which only arises for $m\ge 3$).

Let us start with the first case $K \ge m$ (so that $x$ is small). Consider the function $f_1$ which is the characteristic function of the interval $(-x/(2\pi), x/(2\pi))$.

Its Fourier transform is
$$
\hat{f_1}(\xi) = \int_{-x/(2\pi)}^{x/(2\pi)} e^{-2\pi i t \xi} dt = \frac{\sin (x\xi)}{\pi \xi},
$$
interpreted naturally as $x/\pi$ for $\xi =0$. Put $f_m$ to be the convolution of $f_1$ with itself $m$ times. Then $f_m$ is
supported in $(-mx/(2\pi), mx/(2\pi))$ which is a subset of $(-1/2,1/2)$ (since $x \le \pi/K \le \pi/m$), and ${\hat f}_m(\xi) = {\hat f_1}(\xi)^m$. The Poisson summation formula now
gives
$$
0 = \sum_{k \in {\Bbb Z}} f_m (k+1/2) = \sum_{k \in {\Bbb Z}} (-1)^k {\hat f_m}(k) = \sum_{k\in {\Bbb Z}} (-1)^{k} \Big( \frac{\sin(kx)}{\pi k} \Big)^m,
$$
and rearranging we find that
$$
\sum_{k=1}^{\infty} (-1)^{k-1} \Big( \frac{\sin (kx)}{k} \Big)^{m} = \frac 12 x^m.
$$
Therefore for $n\ge K+1$ (which we may assume) we have
$$
G_m(x;n) \ge \frac {x^m}{2} - \Big| \sum_{k=n+1}^{\infty} (-1)^{k-1}\Big(\frac{\sin (kx)}{k}\Big)^m\Big| \ge \frac{x^m}{2} - \Big(\frac{m}{2(m-1)}\frac{x}{(n+1)^{m-1}}+\frac{1}{(n+1)^m}\Big),
$$
and since $n+1 \ge K+2 \ge \pi/x$, it follows that
$$
G_m(x;n) \ge x^m \Big( \frac 12 - \frac{m}{2(m-1) \pi^{m-1}} - \frac{1}{\pi^m} \Big) > x^m \Big( \frac{1}{2} - \frac{1}{\pi} - \frac{1}{\pi^2} \Big) > 0.
$$
This finishes the first case when $K \ge m$.

Now we turn to the second case $2 \le K < m$, which only happens for $m\ge 3$. As noted earlier, we may suppose that $n\ge K+1$, and write
$$
G_m(x;n) = x^m \sum_{k=1}^{K} (-1)^{k-1} \Big( \frac{\sin (kx)}{k x}\Big)^{m} + \sum_{k=K+1}^{n} (-1)^{k-1} \Big( \frac{\sin (kx)}{k}\Big)^m.
$$
As noted earlier $(\sin y)/y$ is non-negative and decreasing on $[0,\pi)$, and therefore by pairing consecutive terms in the first sum (and if there is a last term
left unpaired it is non-negative) and discarding all but the first two terms, we see that the first sum above is
$$
\ge (\sin x)^m - \Big(\frac{\sin 2x}{2} \Big)^m = (\sin x)^m (1- (\cos x)^m) \ge (\sin x)^{m} (\sin x)^2 \ge \Big( \frac{2}{\pi} x\Big)^m \Big( \frac{2}{K+1} \Big)^2,
$$
where we used that $\sin t \ge 2t/\pi$ for $t\in [0, \pi/2]$. As for the second sum, using our inequality for tails this is bounded in size by
$$
\frac{m}{2(m-1)} \frac{x}{(K+1)^{m-1} } + \frac{1}{(K+1)^{m-1}} \le x^m \Big( \frac{3}{4 \pi^{m-1}} + \frac{1}{\pi^m} \Big).
$$
Since $K<m$, it follows that
$$
G_m(x;n) > x^m \Big( \Big(\frac{2}{\pi}\Big)^m \frac{4}{m^2} - \frac{3}{4 \pi^{m-1}} - \frac{1}{\pi^m} \Big) >0,
$$
since $m\ge 3$. This completes the proof of the second case.

Lastly it remains to verify the estimate for tails. This is proved by partial summation. The desired sum is
$$
x^m \sum_{k=A}^{B} (-1)^{k-1} \Big(\frac{\sin (kx)}{kx} \Big)^m = x^m \int_{A^-}^{B^+} \Big( \frac{\sin (yx)}{yx} \Big)^{m} d\Big( \sum_{A \le k < y} (-1)^{k-1} \Big),
$$
and integrating by parts (and note that $|\sum_{A \le k < y} (-1)^{k-1} | \le 1$ always) we can bound this in magnitude by
$$
x^m \Big| \frac{\sin (Bx)}{Bx} \Big|^m + x^m \int_A^B \Big| \frac{d}{dy} \Big( \frac{\sin (yx)}{yx} \Big)^m \Big| dy.
$$
Now
\begin{align*}
\Big| \frac{d}{dy} \Big( \frac{\sin (yx)}{yx}\Big)^m \Big| &= m \Big| \frac{\sin (yx)}{yx} \Big|^{m-1} \Big| \frac{\cos (yx)}{y} - \frac{sin (yx)}{ y^2x} \Big|
\\
& \le m \Big| \frac{(\sin yx)^{m-1} (\cos yx)}{y^m x^{m-1}} \Big| + \frac{m}{y^{m+1} x^m} \le \frac{m}{2 y^mx^{m-1} } +\frac{m}{y^{m+1} x^m},
\end{align*}
where in the last step we used that $|(\sin t)^{m-1} (\cos t)| \le |\sin t \cos t| \le 1/2$. Using this we conclude that our desired sum is
$$
\le \frac{1}{B^m} + \int_A^B \Big( \frac{mx}{2y^m} + \frac{m}{y^{m+1}} \Big) dy < \frac{mx}{2(m-1) A^{m-1}} + \frac{1}{A^m},
$$
which completes the proof.