x^16-1の因数分解

「x^8+1 の因数分解」で検索された．＞x^n-1の因数分解
ここには，$x^8-1$ までしか書いていないが，$x^{16}-1$ も同様である．

$x^{16}-1\\
=(x-1)(x+1)(x^2+1)(x^4+1)(x^8+1)$

この各因数を因数分解すると，
$x^2+1=x^2-(-1)=x^2-i^2=(x-i)(x+i)$

$x^4+1\\
=x^4+2x^2+1-2x^2\\
=(x^2+1)^2-(\sqrt{2}x)^2\\
=(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)\\
=(x-(\frac{\sqrt{2}+i\sqrt{2}}{2}))\\
(x-(\frac{\sqrt{2}-i\sqrt{2}}{2}))\\
(x-(\frac{-\sqrt{2}+i\sqrt{2}}{2}))\\
(x-(\frac{-\sqrt{2}-i\sqrt{2}}{2}))$

$x^8+1\\
=x^8+2x^4+1-2x^4\\
=(x^4+1)^2-(\sqrt{2}x^2)^2\\
=(x^4-\sqrt{2}x^2+1)(x^4+\sqrt{2}x^2+1)$

前半の4次式は
$x^4-\sqrt{2}x^2+1\\
=x^4+2x^2+1-2x^2-\sqrt{2}x^2\\
=x^4+2x^2+1-(2+\sqrt{2})x^2\\
=(x^2+1)^2-(\sqrt{2+\sqrt{2}}x)^2\\
=(x^2-\sqrt{2+\sqrt{2}}x+1)(x^2+\sqrt{2+\sqrt{2}}x+1)$

前半の2次式は
$x^2-\sqrt{2+\sqrt{2}}x+1\\
=(x-(\frac{\sqrt{2+\sqrt{2}}}{2}+i\frac{\sqrt{2+\sqrt{2}}}{2}))
(x-(\frac{\sqrt{2+\sqrt{2}}}{2}-i\frac{\sqrt{2+\sqrt{2}}}{2}))$

他の項も同様に変形すると，
$x^{16}-1\\
=(x^8-1)(x^8+1)\\
=(x^4-1)(x^4+1)(x^4-\sqrt{2}x^2+1)(x^4+\sqrt{2}x^2+1)\\
=(x^2-1)(x^2+1)(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)(x^2+\sqrt{2+\sqrt{2}}x+1)(x^2-\sqrt{2+\sqrt{2}}x+1)(x^2+\sqrt{2-\sqrt{2}}x+1)(x^2-\sqrt{2-\sqrt{2}}x+1)\\
=(x-1)(x+1)(x-i)(x+i)\\
(x-(\frac{\sqrt{2}+i\sqrt{2}}{2}))\\
(x-(\frac{\sqrt{2}-i\sqrt{2}}{2}))\\
(x-(\frac{-\sqrt{2}+i\sqrt{2}}{2}))\\
(x-(\frac{-\sqrt{2}-i\sqrt{2}}{2}))\\
(x-(\frac{\sqrt{2+\sqrt{2}}}{2}+i\frac{\sqrt{2+\sqrt{2}}}{2}))\\
(x-(\frac{\sqrt{2+\sqrt{2}}}{2}-i\frac{\sqrt{2+\sqrt{2}}}{2}))\\
(x-(-\frac{\sqrt{2+\sqrt{2}}}{2}+i\frac{\sqrt{2+\sqrt{2}}}{2}))\\
(x-(-\frac{\sqrt{2+\sqrt{2}}}{2}-i\frac{\sqrt{2+\sqrt{2}}}{2}))\\
(x-(\frac{\sqrt{2-\sqrt{2}}}{2}+i\frac{\sqrt{2-\sqrt{2}}}{2}))\\
(x-(\frac{\sqrt{2-\sqrt{2}}}{2}-i\frac{\sqrt{2-\sqrt{2}}}{2}))\\
(x-(-\frac{\sqrt{2-\sqrt{2}}}{2}+i\frac{\sqrt{2-\sqrt{2}}}{2}))\\
(x-(-\frac{\sqrt{2-\sqrt{2}}}{2}-i\frac{\sqrt{2-\sqrt{2}}}{2}))$


以上が，複2次式の式変形による因数分解であるが，$x^{16}-1=0$の解は，1の16乗根であるので，複素平面上の原点を中心とする単位円周上，(1,0)を頂点とする正16角形の頂点に並ぶ．＞1の累乗根が円周上にある理由
つまり， k=0,1,2,3,…,15のとき，
$\cos\frac{k\pi}{8}+i\sin\frac{k\pi}{8}$
と表される．
たとえば，k=1 のときは，
$x=\cos\frac{\pi}{8}+i\sin\frac{\pi}{8}$
なので，π/8 の cos と sin を求めればよい．

半角公式で，
$(\sin\frac{\pi}{8})^2=\frac{1-\cos\frac{\pi}{4}}{2}\\
=\frac{1-\frac{1}{\sqrt{2}}}{2}\\
=\frac{2-\sqrt{2}}{4}$
$(\cos\frac{\pi}{8})^2=\frac{1+\cos\frac{\pi}{4}}{2}\\
=\frac{1+\frac{1}{\sqrt{2}}}{2}\\
=\frac{2+\sqrt{2}}{4}$
より，
$x=\cos\frac{\pi}{8}+i\sin\frac{\pi}{8}\\
=\sqrt{\frac{2+\sqrt{2}}{4}}+i\sqrt{\frac{2-\sqrt{2}}{4}}\\
=\frac{\sqrt{2+\sqrt{2}}}{2}+i\frac{\sqrt{2-\sqrt{2}}}{2}$
を得るので，因数の一つは，
$(x-(\frac{\sqrt{2+\sqrt{2}}}{2}+i\frac{\sqrt{2-\sqrt{2}}}{2}))$
とわかる．
あとは，
π/4，3π/8，π/2，5π/8，3π/4，7π/8，π，9π/8，5π/4，11π/8，3π/2，13π/8，7π/4，15π/8
のときの sin cos を求めればよい．

この中で，x^2-1=0，x^4-1=0，x^8-1=0 はそれぞれ，1の平方根，4乗根，8乗根なので，1の原始16乗根（つまり x^8+1=0 の解）は
π/8，3π/8，5π/8，7π/8，9π/8，11π/8，13π/8，15π/8
の sin cos を求めればよい．