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[Jonathan]

If you want to be such a pedant, a sine wave doesn't necessarily have an average of zero. It can be offset from zero. You've just assumed that all sine waves are drawn with zero as the base line.

That's just changing your reference point - if it's got a DC offset then it's a sine wave with an offset, not a sine wave. I thought it was misleading to just say 'squrt(2) is essentially just the average value of a sine wave'.

As I'm sure you're aware, I meant that the average value for 90degree either side of a sine wave peak or trough i.e a 180degree section, is sqrt(2).

No that's a new one to me, as it's incorrect!
As I'm sure you're aware, to find the average value of a function you integrate it over the section in question, and divide by the 'length' of that section. So lets do it:
Let y=sin(x)
We want '90degree[sic] either side of a sine wave peak', so in radians the limits of out integral are 0 and pi and we integrate over pi:
Average=1/pi*Integral(sin(x)dx) from 0 to pi.
Average=1/pi*(-cos(pi)--cos(0))
Average=2/pi

2/pi is not equal to sqrt(2).

To get the correct RMS value of a function, you first square the function, then find the mean of the function (using the above method) then find the square root of it. Hence the name - root mean square.