I am trying to understand how to obtain the Fourier transform of the Bartlett (triangular) window. The Bartlett window is defined as $$ w_B(k)=\begin{cases}\frac{N-|k|}N,& |k|\le N;\\0,&|k|>N.\end{cases} $$ The Fourier transform of the Bartlett window is then given by $ \sum_{k=-\infty}^\infty w_B(k)e^{-ik\omega}$.
Using the expression of the Bartlett window and the fact that $e^{jk\omega}+e^{-jk\omega}=2\cos(k\omega)$, $$ \sum_{k=-\infty}^\infty w_B(k)e^{-ik\omega}=1+2\sum_{k=1}^N\frac{N-k}N\cos(k\omega). $$ Since $$ \sum_{k=0}^N\cos(k\omega) =\frac12+\frac{\sin((N+1/2)\omega)}{2\sin(\omega/2)}, $$ $$ \sum_{k=1}^Nk\cos(k\omega) =\frac{(N+1)\sin((N+1/2)\omega)}{2\sin(\omega/2)}-\frac{1-\cos((N+1)\omega)}{4\sin^2(\omega/2)} $$ and $$ \frac{1-\cos((N+1)\omega)}2 =\sin^2((N+1)\omega/2), $$ we obtain $$ \tag{#} \sum_{k=-\infty}^\infty w_B(k)e^{-ik\omega}=\frac1N\biggl[\frac{\sin^2(N\omega/2+\omega/2)}{\sin^2(\omega/2)}-\frac{\sin(N\omega+\omega/2)}{\sin(\omega/2)}\biggr]. $$ However, the answer should be $$ \sum_{k=-\infty}^\infty w_B(k)e^{-ik\omega}=\frac1N\biggl[\frac{\sin(N\omega/2)}{\sin(\omega/2)}\biggr]^2. $$ How can we obtain the answer from $(\#)$? Is there a simpler way to obtain the Fourier transform of the Bartlett window?
Any help is much appreciated!
The desired equality is $$\frac1N\biggl[\frac{\sin^2(N\omega/2+\omega/2)}{\sin^2(\omega/2)}-\frac{\sin(N\omega+\omega/2)}{\sin(\omega/2)}\biggr]=\frac1N\biggl[\frac{\sin(N\omega/2)}{\sin(\omega/2)}\biggr]^2,$$ which upon multiplying both sides by $N\sin(\omega/2)^2$ and rearranging a bit simplifies to
$$\sin(\omega/2)\sin(N\omega+\omega/2)=\sin^2(N\omega/2+\omega/2)-\sin(N\omega/2)^2.$$ But if we recall the product-to-sum trigonometric formulas, we can write the LHS as $$\frac{1}{2}\biggl[\cos(N\omega) -\cos(N\omega+\omega) \biggr],$$ which we recognize as the RHS upon two applications of the double-angle formula $\cos 2\theta=1-2\sin^2\theta$. Hence the announced equality is shown.