Consider the sequence
$$ r[n] = \left\{ \begin{array}{ll} 1, & 0\leq n\leq M \\ 0, & \text{otherwise} \\ \end{array} \right. $$
for which
$$ R(e^{j\omega}) = e^{-j\frac{M}{2}\omega}\times\frac{ sin(\frac{M+1}{2} \omega) } {sin(\frac{\omega}{2})} $$
Find the Fourier transform of the following sequence in terms of $R(e^{j\omega})$:
$$ w[n] = \left\{ \begin{array} (\frac{1}{2}[1 - cos(\frac{2\pi n}{M})], & 0\leq n\leq M \\ 0, & \text{otherwise} \\ \end{array} \right. $$
Solution that I think is wrong:
Since, $$ w[n] = r[n] \cdot \frac{1}{2}[1+cos(\frac{2\pi n}{M})] $$
$$ W(e^{j\omega}) = R(e^{j\omega})\sum_{n=-\infty}^{\infty}\frac{1}{2}(1+cos(2\pi n/M))(e^{-j\omega n}) \\ = R(e^{j\omega})\sum_{n=-\infty}^{\infty}(\frac{1}{2}+\frac{1}{4}e^{j2\pi n/M}+\frac{1}{4}e^{-j2\pi n/M})e^{-j\omega n} \\ = R(e^{j\omega})\sum_{n=-\infty}^{\infty}(\frac{1}{2}\delta(\omega)+\frac{1}{4}\delta(\omega+2\pi /M)+\frac{1}{4}\delta(\omega-2\pi /M)) \\ $$
I think this solution is wrong because the Fourier transform of the product of two sequence is not the product of their respective Fourier transforms. Rather, it is the periodic convolution of their respective Fourier transforms.
Moreover, the sum $\sum_{n=-\infty}^{\infty}e^{-j\omega n}$ does not converge. The Fourier transform for $x[n]=1$ is not derived from that sum, and also is $2\pi \delta(\omega)$ not $\delta(\omega)$ as used above.
My solution:
From the modulation/windowing theorem,
$$ W(e^{j\omega}) = \frac{1}{2\pi}\int_{-\pi}^{\pi}R(e^{j\omega})\cdot 2\pi(\frac{1}{2}\delta(\omega - \theta)+\frac{1}{4}\delta(\omega-\theta+\frac{2\pi}{M})+\frac{1}{4}\delta(\omega - \theta\ + \frac{2\pi}{M})) d\theta \\ = \frac{1}{2}R(e^{j\omega}) + \frac{1}{4}R(e^{j(\omega-2\pi/M)})+\frac{1}{4}R(e^{j(\omega+2\pi/M)}) $$
Is my reasoning for why the first solution is wrong, correct? Is my solution correct?