Real and imaginary part of a complex sinusoid $y(t)=\sin(wt)$ I'm trying to understand the plots on this page. It's a book about the Discrete Fourier Transform and it's discussing how a a function $x(t)=\cos(w_0t)$ or $y(t)=\sin(w_0t)$ is composed of a positive and a negative frequency component. I get why the spectrum of $\cos(wt)$ has two real components and none imaginary. But i don't get why $\sin(wt)$ have two imaginary components, as in b) of the following image.
This is the link for the image, from the web mentioned page, that i don't understand
I think i get how $x(t)=\cos(wt)$ is the sum of two complex sinusoids of frequencies of opposite signs that results in an zero imaginary part:
$$x(t)=\cos(wt)=\frac{e^{jwt}+e^{-jwt}}{2}$$
$$  x(t)=\frac{\cos(wt)+j\sin(wt)+\cos(-wt)+j\sin(-wt)}{2}  $$
Since 
$$\cos(-x)=\cos(x)$$
$$\sin(-x)=-\sin(x)$$ 
follows
$$  x(t)=\frac{\cos(wt)+j\sin(wt)+\cos(wt)-j\sin(wt)}{2}  $$
so
$$Re\{ \ x(t)\ \} = \frac{\cos(wt)+\cos(wt)}{2}=\cos(wt)$$
and
$$Im\{ \ x(t)\ \} = \frac{\sin(wt)-\sin(wt)}{2}=0$$
That explains why $\cos(wt)$ have two real parts on the graph, of same amplitude and "opposite" frequencies.
I will try to to the same with $\sin(wt)$:
$$y(t)=\sin(wt)=\frac{e^{jwt}-e^{-jwt}}{2j}$$
Using $\cos(-x)=\cos(x)$  
$$y(t)=\frac{ \cos(wt)+j\sin(wt) -(\cos(wt)+j\sin(-wt)) }{  2j  }$$
$$y(t)=\frac{ \cos(wt)+j\sin(wt) -\cos(wt)-j\sin(-wt) }{  2j  }$$
$$Re\{ \ y(t) \ \}=\frac{\cos(wt)-\cos(wt)}{2j}=0$$
$$Im\{ \ y(t) \ \}=\frac{\sin(wt)-\sin(-wt)}{2j}$$
I'm not sure how to follow from there. How come an imaginary part contains $j$? Or maybe $j$ should not be included? But in the case of
$$Im\{ \ y(t) \ \}=\frac{\sin(wt)-\sin(-wt)}{2}$$
Where that $j^-1$ went? This looks wrong to me because 
$$j \cdot Im\{ \ y(t) \ \}\neq\frac{\sin(wt)-\sin(-wt)}{2j}$$
What did I do wrong here? This looks so silly, I'm sorry.
 A: I think that the answer is less complex than you are making it.
Once you have:
$$\cos(\omega t)=\frac{e^{j \omega t}+e^{j (-\omega) t}}{2} $$
that shows that there is a '$\frac{1}{2}$' magnitude at '$\omega$' and a '$\frac{1}{2}$' magnitude at '$-\omega$', both in the positive real direction.
Taking the same logic:
$$\sin(\omega t)=\frac{e^{j \omega t}-e^{j (-\omega) t}}{2j} $$
Multiply both numerator and denominator of the fraction by $\frac{j}{2}$:
$$\sin(\omega t)=\frac{j\frac{1}{2} e^{j \omega t} - j\frac{1}{2} e^{j (-\omega) t}}{\frac{1}{2}\times 2j^2}$$
Where $j^2=-1$. This eliminates the denominator and multiplies the numerator by $-1$:
$$\sin(\omega t)=-j\frac{1}{2} e^{j \omega t} + j\frac{1}{2} e^{j (-\omega) t}$$
This shows that there is a '$-\frac{1}{2}j$' point at '$\omega$' and a '$(+)\frac{1}{2}j$' point at '$-\omega$'. In this case, both are in the imaginary plane. The one at positive $\omega$ has a negative sense, and the one at negative $\omega$ has a positive sense.
Does that help?
