倍角・３倍角・半角の公式の導出

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倍角

加法定理より

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\begin{array}{ll}

\sin 2x &= \sin(x+x)\\
&= \sin x \cos x + \cos x \sin x\\
&= 2 \sin x \cos x\\
\\
\cos 2x &= \cos(x + x)\\
&= \cos x \cos x - \sin x \sin x\\
&= \cos^2 x - \sin^2 x\\
\\
&= \cos^2 x - (1 - \cos^2 x) = 2 \cos^2 x - 1\\
&= (1 - \sin^2 x) - \sin^2 x = 1- 2 \sin^2 x\\
\\
\tan 2x &= \tan(x + x)\\
&= \frac{\tan x + \tan x}{1 - \tan x \tan x}\\
&= \frac{2 \tan x}{1 - \tan^2 x}\\
\end{array}

３倍角

加法定理と倍角の公式より

\\
\begin{array}{ll}

\sin 3x &= \sin(2x + x)\\
&= \sin 2x \cos x + \cos 2x \sin x\\
&= (2 \sin x \cos x) \cdot \cos x + (1 - 2 \sin^2 x) \cdot \sin x\\
&= 2 \sin x \cos^2 x + \sin x - 2 \sin^3 x\\
&= 2 \sin x (1 - \sin^2 x) + \sin x - 2 \sin^3 x\\
&= -4 \sin^3 x + 3 \sin x\\
\\
\cos 3x &= \cos(2x + x)\\
&= \cos 2x \cos x - \sin 2x \sin x\\
&= (2 \cos^2 - 1) \cdot \cos x - (2 \sin x \cos x) \cdot \sin x\\
&= 2 \cos^3 x - \cos x - 2 \sin^2 x \cos x\\
&= 2 \cos^3 x - \cos x - 2 (1 - \cos^2 x) \cos x\\
&= 4 \cos^3 x - 3 \cos x\\
\end{array}

半角

倍角の公式より

\\
\begin{array}{ll}
\sin 2x &= 1 - 2 \sin^2 x\\
2 \sin^2 x &= 1 - \sin 2x\\
\sin^2 x &= \frac{1 - \sin 2x}{2}\\
\\
\cos 2x &= 2 \cos^2 x - 1\\
2 \cos^2 x &= 1 + \cos 2x\\
\cos^2 x &= \frac{1 + \cos 2x}{2}\\
\\
\tan^2 x &= \frac{\sin^2 x}{\cos^2 x}\\
&= \frac{\frac{1 - \sin 2x}{2}}{\frac{1 + \cos 2x}{2}}\\
&= \frac{1 - \sin 2x}{1 + \cos 2x}\\
\end{array}