ヴィエトの等式

\begin{align}
&\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{2+\sqrt{2}}}{2}\cdot\frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2}\cdot\dots\\
&=\frac{2}{\pi}\\
\\
\sin x&=2\cos\frac{x}{2}\sin\frac{x}{2}\\
&\sin\frac{x}{2}=2\cos{x}{4}\sin\frac{x}{4}なので，\\
&=2\cos\frac{x}{2}\cdot2\cos\frac{x}{2^2}\sin\frac{x}{2^2}\\

\frac{\sin x}{x}&=2^n\cos\frac{x}{2}\cos\frac{x}{2^2}\dots\dots\cos\frac{x}{2^n}\cdot\frac{\sin\frac{x}{2^n}}{\frac{x}{2^n}}\\
n\to &\inftyのとき，\frac{\sin\frac{x}{2^n}}{\frac{x}{2^n}}\to1\\
した&がって，\frac{\sin x}{x}を表す無限級数は，\\
\frac{\sin x}{x}&=\cos\frac{x}{2}\cos\frac{x}{2^2}\cos\frac{x}{2^3}\dots\dots\\
x=&\frac{\pi}{2}のとき，\\
\frac{\sin\frac{\pi}{2}}{\frac{\pi}{2}}&=\frac{2}{\pi}=\cos\frac{\pi}{4}\cos\frac{\pi}{8}\cos\frac{\pi}{16}\dots\dots\\
{\cos}^2\frac{\theta}{2}&=\frac{1+\cos\theta}{2}より，\\
0\leqq\theta&\leqq\piのとき，\cos\frac{\theta}{2}=\frac{\sqrt{2+2\cos\theta}}{2}\\
\cos\frac{\pi}{4}&=\frac{\sqrt{2}}{2}\\
\cos\frac{\pi}{8}&=\frac{\sqrt{2+\sqrt{2}}}{2}\\
\cos\frac{\pi}{16}&=\frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2}\\
\dots\dots\\
ゆえに，&\frac{2}{\pi}=\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{2+\sqrt{2}}}{2}\cdot\frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2}\cdot\dots
\end{align}