<s>[INST] Why is one root extraneous when using complex number polynomial expansions of cos(4x) to find cos(π/8)? Using De Moivre's theorem, binomial expansion, and the Pythagorean identity, I have the following polynomial for $\cos 4\theta$:
$$\cos\left(4\theta\right) = 8\cos^4\theta -8\cos^2\theta +1 $$
I trying to find an exact value for $\cos\left(\frac{\pi}{8}\right)$.
Since $\cos\left(\frac{\pi}{2}\right)=0$, then $\cos\left(4\cdot\frac{\pi}{8}\right)=0$, so it must be true that
$$ 8\cos^4\left(\frac{\pi}{8}\right) -8\cos^2\left(\frac{\pi}{8}\right) +1 =0$$
and by the quadratic formula I have
$$ \cos \left(\frac{\pi}{8}\right) = \pm\frac{\sqrt{2\pm\sqrt{2}}}{2} $$
Since $\frac{\pi}{8}$ is in the first quadrant, we discard the negative root, so we have
$$  \cos \left(\frac{\pi}{8}\right) = \frac{\sqrt{2\pm\sqrt{2}}}{2} $$
Now, $  \cos \left(\frac{\pi}{8}\right) = \frac{\sqrt{2+\sqrt{2}}}{2} $ is true, but  $  \cos \left(\frac{\pi}{8}\right) = \frac{\sqrt{2-\sqrt{2}}}{2} $ is false. (I know this from the approximate numerical value of $\cos\left(\frac{\pi}{8}\right)$.)
My question: How do we know $\frac{\sqrt{2-\sqrt{2}}}{2}$ is an extraneous root?
 [/INST] Since $0<\frac\pi8<\frac\pi4$, and since $\cos|_{[0,\pi/2]}$ is strictly decreasing,$$1>\cos\left(\frac\pi8\right)>\cos\left(\frac\pi4\right)=\frac{\sqrt2}2.$$But$$\frac{\sqrt{2-\sqrt2}}2<\frac{\sqrt2}2.$$
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