Explaining why sin and cos are *not* at right angles Disclaimer: I'm an engineer, not a mathematician
I recently had a fierce discussion (lots of blood) on electronics.stackexchange about phase shifts.  
The impedance of a resistor is real, that of a coil imaginary ($j\omega L$), so that adding 3V across a resistor and 4V across a coil in series results in a 5V overall.  
Now John (let's call him John) explained the same by using a sine and a cosine function, claiming they're at 90°. I can see where he gets this, and I tried the following to explain that the sine of a number and the cosine of a number are scalars, not vectors, so they can't have a phase difference:
$$\sin(\omega t) = \dfrac{e^{j \omega t} - e^{- j \omega t}}{2j}$$
This is easy for me to visualize: $e^{\omega t}$ and $e^{- \omega t}$ are phasors rotating in opposite directions in the complex plane. Their difference is a vector on the imaginary axis. Dividing by $j$ rotates that vector by $\pi /2$ clockwise, so that it moves from the imaginary axis to the real axis. And then it's a sine, a scalar. Sitting nicely next to the cosine, no phase difference. 
My problem is that half a second ago it was still a vector. How does it become a scalar? Or do "being a vector on the real axis" and "being a scalar" mean the same thing?
 A: I'm a mathematician, not an engineer, so this may not be the "right" answer to your question, but hopefully will give some insight into the mathematical language (and expands on fishlips' comment).
Typically, the word "vector" means an element of a vector space, for example $\mathbb{R}^3$, or the set of functions $\mathbb{R}\to\mathbb{R}$, and the word "scalar" means an element of a field, such as $\mathbb{R}$ or $\mathbb{C}$. The problem with this terminology is that fields are themselves vector spaces, so depending on your point of view an element of them can either be a vector (if you're thinking about the vector space structure) or a scalar (if you're thinking about the field structure).
The field $\mathbb{C}$ as in your example can be particularly complicated in this respect, as it is a field, a one-dimensional complex vector space, and a two-dimensional real vector space. My guess is it's the last interpretation that is most useful to you, so you should interpret $\sin(\omega t)$ as a vector that happens to lie in the real axis, and not necessarily as a scalar. (But maybe that's not the interpretation you want, hence the disclaimer at the beginning).
Zhen Lin's comment is also worth expanding on - as I mentioned above, sets of functions can be vector spaces. As $\sin$ and $\cos$ are both functions $\mathbb{R}\to\mathbb{R}$, they are vectors in the vector space consisting of all such functions, and after choosing an appropriate inner product (the $L^2$ inner product), these vectors are orthogonal to each other. (If you don't have the time or inclination to read through the inner product article, the relevant feature of inner products  is that they tell you what it means for two vectors to be orthogonal.)
A: To comment on this:
I can see where he gets this, and I tried the following
to explain that sine and cosine are scalars, not vectors,
so they can't have a phase difference:

This is wrong. First off, the notions of 'scalar' and 'vector' aren't really relevant here (directly, anyways). The basic definition of phase is that it's a property of certain real-valued functions of the real numbers. Specifically, for any function $f$ that can be written as
$$ f(t) = B + A \sin(\omega t + \phi) $$
then we define $\mathop{\mathrm{Phase}}(f) = \phi$. Phase is an "angular position" meaning that it's only modulo $2 \pi$ radians (or modulo 360 degrees, if you prefer angles to be measured that way).
As a point of grammar, the type of "sin" is "function", and the type of "sin 3" is "number". The type of "sin t" should also be a "number" (the number one gets by plugging "t" into "sin") -- but by a common abuse of notation is also used to denote the function "sin".

Now that I have the link, I think the meat of John's argument is as follows.
The main point is the sum-of-sines identity which is, if I haven't made an error along the way,
$$ A \sin(\omega t) + B\sin(\omega t + \phi)
= C \sin(\omega t + \theta) $$
where
$$ C^2 = (A + B \cos \phi)^2 + (B \sin \phi)^2 $$
$$ \tan(\theta) = (B \sin \phi) / (A + B \cos \phi) $$.
The main point you can see is that if the relative phase between the two sine waves is given by $\phi = \pi/2$, then the amplitude of the sum of the two sine waves is given by
$$ C^2 = A^2 + B^2 $$
My knowledge of electrical engineering is limited and rusty -- but I believe the point is that the current across the resistor and the coil are indeed sine waves 90 degrees out of phase, and that you seek to add them.

One interesting thing to note is that this gives a justification for representing a sine wave
$$A \sin(\omega t + \phi)$$
by the plane vector
$$\langle A \cos \phi, A \sin \phi \rangle$$
because the formula shows that adding sine waves corresponds to adding vectors. Additionally, the magnitude of the wave is the magnitude of the vector, and the phase of the wave is the polar angle of the vector.
I know I knew this once upon a time, but I had long forgotten. It's still a neat fact, I think. :)
(A more boring, but somewhat more obvious statement of this same fact is that $\{ \cos \omega t, \sin \omega t \}$ are a basis for the set of sine waves with period $2 \pi / \omega$)
A: If you decide to view $e^{\omega t}$ as a vector (which is a useful point of view, but not the only one possible), then you should also interpret $j$ as a vector (namely, $(0,1)$).
Now, your fraction is the proportion of two parallel vectors which is a scalar.

On the other hand, if you prefer to interpret $j$ as rotation acting on the space, then you should also interpret $e^{\omega t}$ as rotation on the space. In that case, your expression turns out to be the function that is the scalar multiplication by $\sin$.
