【数Ⅲ複素数平面】ド・モアブルの定理を使って3倍角の公式を導出する
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ド・モアブルの定理を利用した三倍角の公式の導出
では実際に複素平面の極形式の基本的な形$\cos\theta+i \sin\theta$を使って3倍角の公式を求めてみましょう。
ド・モアブルの定理より
$\cos3\theta+i\sin3\theta=(\cos\theta+i \sin\theta)^3$
$=\cos^3\theta+3i\cos^2\theta\sin\theta-3\cos\theta\sin^2\theta-i\sin^3\theta$
$=(\cos^3\theta-3\cos\theta\sin^2\theta)+i(3\cos^2\theta\sin\theta-\sin^3\theta)$
複素数における恒等式の性質を利用する
ここで恒等式の性質を利用して、左右の実部と虚部をそれぞれ比較していきます。複素数では[実部]=[実部]、[虚部]=[虚部]の関係が成り立ちます。例えば、$a+bi=3+5i$なら$a=3,b=5$であることが分かります。
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では、実際に左右の実部と虚部でそれぞれ恒等式を作っていきます。まずは、実部どうしで式を作ります。
$\cos^3\theta-3\cos\theta\sin^2\theta=\cos 3\theta$
$\cos3\theta=\cos^3\theta-3\cos\theta(1-\cos^2\theta)$
$=\cos^3\theta-3\cos\theta+3\sin^3\theta$
$=4\cos^3\theta-3\cos\theta$
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また、左右の虚部を比較すると
$\sin3\theta=3\cos^2\theta\sin\theta-\sin^3\theta$
$=3(1-\sin^2\theta)\sin\theta-\sin^3\theta$
$=3\sin\theta-3\sin^3\theta-\sin^3\theta$
$=3\sin\theta-4\sin^3\theta$
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