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 $x/(K+1) < x\le \pi/K$.  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.