I'm told to find the following signal as just a sum of cosines. The only Fourier coefficients that aren't zero are
Where $X[k]$ comes from $x(t) = \sum_{k = -\infty}^{\infty}X[k]e^{i \omega tk}$ and $j = \sqrt{-1} = i$.
My approach: We want to find $x(t)$ as a sum of only cosines. We're told that the period is $T = 8$, so $\omega = \frac{\pi}{4}$
I'm taking the $^{*}$ in $X^{*}[-3]$ to mean complex conjugate... (hopefully) ... please tell me if I'm wrong. So $X[k = -3] = -4i$
$$x(t) = X[k = -3]e^{i\frac{-3\pi}{4}t} + X[k = -1]e^{i\frac{-\pi}{4}t} + X[k = 1]e^{i\frac{\pi}{4}t} + X[k = 3]e^{i\frac{3\pi}{4}t}$$
Now substituting everything in
$$x(t) = -4ie^{i\frac{-3\pi}{4}t} + 2e^{i\frac{-\pi}{4}t} + 2e^{i\frac{\pi}{4}t} + 4ie^{i\frac{3\pi}{4}t}$$ $$x(t) = 4\cos(\frac{\pi}{4}t) + (4i)(e^{i\frac{3\pi}{4}t} - e^{i\frac{-3\pi}{4}t})$$ $$x(t) = 4\cos(\frac{\pi}{4}t) + (8i^2)(\sin(\frac{3\pi}{4}t))$$ $$x(t) = 4\cos(\frac{\pi}{4}t) - (8)(\cos(\frac{3\pi}{4}t + \frac{\pi}{2}))$$
However the answer is actually,
Does that mean I got the definition of $X^{*}[-3]$ wrong? What does it actually mean? If I don't take the complex conjugate of $4j$ then I have $8j\cos(\frac{3\pi}{4}t)$ which is also incorrect?