# 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?

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They are orthogonal, in the sense of the $L^2$ inner product. (This is why we can decompose functions into Fourier series!) – Zhen Lin May 24 '12 at 10:26
A vector is (by definition) an element of a vector space. $R$ forms a vector space over itself, so the distinction that something is either scalar or vector is not exactly valid. – fishlips May 24 '12 at 10:29
The sine of a number is a number and can't have a phase difference with another number. The sine as a function over the real line is another matter. The sine function and the cosine function obviously have a $90^\circ$ phase difference, or I don't know what phase difference is. – Rahul May 24 '12 at 10:31
I think the comments show that there is some ambiguity in the meaning of "at right angles" - but also that orthogonality is a rich mathematical concept with many applications. The naïve thought that - in a right-angled triangle with hypotenuse 1, sin and cos are literally at right angles is related, as also is the fact that the projections of points on the unit circle onto orthogonal axes track sin and cos at right angles. – Mark Bennet May 24 '12 at 11:14
@user32112, you have to distinguish between the sine of a particular number and the sine function itself. Your argument is kind of like saying, it's not true that the President of the United States is elected every four years, because Barack Obama is the President of the United States, and Barack Obama is not elected every four years. – Rahul May 24 '12 at 11:20

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.)

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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 $2 \pi / \omega$)

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You're right, the sine and cosine in the quote seem to refer to functions. I'll fix it. – stevenvh May 24 '12 at 11:28
The "boring statement" is interesting. I just posted about the Fourier series on electronics.stackexchange, and used this to explain an alternative definition (sum of only sines instead of sin + cos)! :-) – stevenvh May 24 '12 at 12:14

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 $\sin$.

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