*Warning:* trigonometry ahead.

Last time we looked at how to downmix a stereo signal to mono (*M* = *L*/2 + *R*/2). The resulting *M* signal can only be *as powerful* as the *L* and *R* signals if they are *perfectly correlated* (*L* = *R*); if the *L* and *R* signals are *uncorrelated* (no relationship between* L* and *R*), the resulting *M* signal is only *half* as powerful; and if the *L* and *R* signals are* anticorrelated* (*L* = -*R*), the resulting *M* signal vanishes.

This time I want to look more closely at what happens to *M* as {*L*, *R*} varies from correlated via uncorrelated to anticorrelated. In particular I'll take *L* and *R* to be full-scale sine waves of equal frequency and offset by a constant phase: {*L*, *R*} = {sin *θ*, sin(*θ* - *φ*)}. We'll see what happens to *I _{M }*/

*I*as a function of

_{L}*φ*.

Recall from before that the intensity of a signal *s*(*t*) over a time interval (*a*, *b*) is defined as

where the dc value is

Our downmixed signal is defined as

So our dc is zero (note that I am taking (*a*, *b*) to be a single period:)

A refresher on the angle-sum identities, which I will use without proof:

- sin(
*A*+*B*) = sin*A*cos*B*+ cos*A*sin*B*- Corollary: sin(
*A*-*B*) = sin*A*cos*B*- cos*A*sin*B*

- Corollary: sin(
- cos(
*A*+*B*) = cos*A*cos*B*- sin*A*sin*B*- Corollary: cos(
*A*-*B*) = cos*A*cos*B*+ sin*A*sin*B*

- Corollary: cos(

We know from before that *I _{L}* =

*I*= sqrt(1/2) = 0.707.... Let's calculate

_{R}*I*as a function of

_{M}*φ*:

Nice and simple.

Are we done? No.

*When you see a fact, try to see it as intuitively as possible* -- Pólya

OK, let's graph this.

That looks suspiciously like a cosine in absolute value. In particular, after messing around with a graphing calculator, it looks like:

Is this in fact a trigonometric identity? Let's see...

Much nicer and simpler. So why didn't our original derivation land us here?

Let's go back to our original graph and shift our frame of reference forward by *φ*/2. Instead of {*L*, *R*} = {sin *θ*, sin(*θ* - *φ*)} we get {*L*, *R*} = {sin(*θ* + *φ*/2), sin(*θ* - *φ*/2)}. Solving for *I _{M}* now gives:

This result is much easier to digest. The sqrt(1/2) is the intensity of the pre-mixed signal; the loss of intensity due to the downmixing, then, is |cos(*φ*/2)|.