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I'd like to refer the following answer:
https://math.stackexchange.com/a/628992/130682
@robjohn claims that:
$$\cos(x):\left[\frac1{\sqrt2},\frac\pi4\right]\mapsto\left[\frac1{\sqrt2},\frac\pi4\right]$$
$\pi\over 4$ is $a_1$ but where does $1\over \sqrt(2)$ came from?
Update:
My actual question is:
Given $a_1 = {\pi \over 4}$, $a_n = \cos(a_{n-1})$
Why does the range of this recurrence is $\left[\frac1{\sqrt2},\frac\pi4\right]$
Thanks.
$\dfrac{1}{\sqrt{2}}$ is $\cos\left(\dfrac{\pi}{4}\right)$
The statement is that if you take the cosine of any value in $\left[\dfrac{1}{\sqrt{2}},\dfrac{\pi}{4}\right]$ the result will be in $\left[\dfrac{1}{\sqrt{2}},\dfrac{\pi}{4}\right]$.
You should be able to verify this. With regards to your specific question, $\cos\left(\dfrac{\pi}{4}\right) = \dfrac{1}{\sqrt{2}}.$
Our first interval will be $\left[\dfrac{1}{\sqrt{2}},\dfrac{\pi}{4}\right]$ and then our second interval will be: $$\left[\cos\left(\dfrac{\pi}{4}\right),\cos\left(\dfrac{1}{\sqrt{2}}\right)\right] \approx \left[.7071067,.760244...\right].$$ Consider that our original interval was: $$\left[\dfrac{1}{\sqrt{2}},\dfrac{\pi}{4}\right] \approx [.7071067, .78539...]$$
So the statement makes sense.
