$1593$ 年，韦达发表了圆周率无穷乘积极限公式，这是第一个可以直接用于计算圆周率到任意精度的古典公式．推导过程如下：因为 \[\begin{split}\cos\dfrac{\alpha}2\cos\dfrac{\alpha}4\cdots\cos\dfrac{\alpha}{2^{n-1}}\cos\dfrac{\alpha}{2^n}&=\dfrac{\cos\dfrac{\alpha}2\cos\dfrac{\alpha}4\cdots\cos\dfrac{\alpha}{2^{n-1}}\cos\dfrac{\alpha}{2^n}\sin\dfrac{\alpha}{2^n}}{\sin\dfrac{\alpha}{2^n}}\\ &=\dfrac{\cos\dfrac{\alpha}2\cos\dfrac{\alpha}4\cdots\cos\dfrac{\alpha}{2^{n-1}}\sin\dfrac{\alpha}{2^{n-1}}}{2\sin\dfrac{\alpha}{2^n}}\\ &=\cdots\\ &=\dfrac{\sin\alpha}{2^n\sin\dfrac{\alpha}{2^n}},\end{split}\]其中 $n\in\mathbb N^{\ast}$ 且当 $n\rightarrow+\infty$ 时，$2^n\sin\dfrac{\alpha}{2^n}\rightarrow\alpha$，所以\[\cos\dfrac{\alpha}2\cos\dfrac{\alpha}4\cdots\cos\dfrac{\alpha}{2^n}\cdots=\dfrac{\sin\alpha}{\alpha}.\]根据以上信息，计算\[\dfrac{\sqrt{2+\sqrt 2}}2\times\dfrac{\sqrt{2+\sqrt{2+\sqrt 2}}}2\times\dfrac{\sqrt{2+\sqrt{2+\sqrt{2+\sqrt 2}}}}2\times\cdots\]的值为（ ）
A．$\dfrac 1{\pi}$
B．$\dfrac{\sqrt 2}{\pi}$
C．$\dfrac 2{\pi}$
D．$\dfrac{2\sqrt 2}{\pi}$
答案 D．
解析 记 $x_1=\dfrac{\sqrt 2}2$，$x_{n+1}=\dfrac{\sqrt{2+2x_n}}2$，则\[x_n=2x_{n+1}^2-1,\]于是题中代数式即\[\cos\dfrac{\pi}8\cdot \cos\dfrac{\pi}{16}\cdots\cos\dfrac{\pi}{2^n}\cdots=\dfrac{\sin\dfrac{\pi}4}{\dfrac{\pi}4}=\dfrac{2\sqrt 2}{\pi}.\]