Unable to obtain correct correct fourier transform for homework Okay so I have this question for my homework which I am asked to find the fourier transform of a signal. However I am unable to obtain the correct answer, the question and my working is shown below.

Question( The answer is the equation of Xp(f) ):



My working(Look from b) onwards):



As can be seen, I am unable to get the answer. But I don't feel that I am wrong.

Thanks.
If $x(t)$ is the (non-periodic) function in the interval $[-2,2]$, then the periodic function $x_p(t)$ is given by

$$x_p(t)=x(t)\star \sum_{n=-\infty}^{\infty}\delta(t-6n)\tag{1}$$

where $\star$ denotes convolution, and, consequently, its Fourier transform is

$$X_p(f)=X(f)\cdot\frac16\sum_{k=-\infty}^{\infty}\delta\left(f-\frac{k}{6}\right)=
\frac16\sum_{k=-\infty}^{\infty}X\left(\frac{k}{6}\right)\delta\left(f-\frac{k}{6}\right)\tag{2}$$

where $X(f)$ is the Fourier transform of $x(t)$. It remains to find $X(f)$. The function $x(t)$ can be written as

$$x(t)=\sin\left(\frac{\pi t}{2}\right)\cdot\left[\text{rect}\left(\frac{t-1}{2}\right)-\text{rect}\left(\frac{t+1}{2}\right)\right]\tag{3}$$

With the Fourier transform pairs

$$\begin{align}\sin\left(\frac{\pi t}{2}\right)&amp;\Longleftrightarrow \frac{1}{2j}\left[\delta\left(f-\frac14\right)-\delta\left(f+\frac14\right)\right]\\\text{rect}\left(\frac{t\pm 1}{2}\right)&amp;\Longleftrightarrow 2e^{\pm j2\pi f}\text{sinc}(2f)\end{align}\tag{4}$$

we obtain from $(3)$

$$\begin{align}X(f)&amp;=\frac{1}{2j}\left[\delta\left(f-\frac14\right)-\delta\left(f+\frac14\right)\right]\star 2\;\text{sinc}(2f)(e^{-j2\pi f}-e^{j2\pi f})\\&amp;=2\left[\delta\left(f+\frac14\right)-\delta\left(f-\frac14\right)\right]\star\text{sinc}(2f)\sin(2\pi f)\\
&amp;=2\left[\text{sinc}\left(2f+\frac12\right)\sin\left(2\pi f+\frac{\pi}{2}\right)-\text{sinc}\left(2f-\frac12\right)\sin\left(2\pi f-\frac{\pi}{2}\right)\right]\end{align}\tag{5}$$

The final expression for $X_p(f)$ results from combining $(5)$ and $(2)$.