Suppose $U_n = \prod_{k=2}^n \cos\left(\frac{\pi}{2^k}\right)$ and $V_n = U_n\cdot\cos\left(\frac{\pi}{2^n}\right)$.

How can I prove that $U_n$ and $V_n$ are adjacent?

What is the limit of $U_n$ and $v_n$?

Note: $W_n = U_n\cdot\sin\left(\frac{\pi}{2^n}\right)$ and $W_n$ is geometrical.

  • $\begingroup$ What does "adjacent series" mean? $\endgroup$ – DonAntonio Nov 5 '12 at 13:38

By the double angle formula, we have that

$$\sin x=2\sin \frac x 2 \cos \frac x 2 $$

Using the double angle formula once more

$$\sin x=2\cdot 2 \sin \frac x 4\cos \frac x 4 \cos \frac x 2 $$

You should realize that by induction,

$$\sin x=2^n \sin \frac x {2^n} \prod_{k=1}^n \cos \frac x {2^k} $$

This means that, for $x\neq 0$.

$$\frac{\sin x}x=\frac{2^n}x \sin \frac x {2^n} \prod_{k=0}^n \cos \frac x {2^k} $$

$$\frac{\sin x}{x}= \frac{\sin \dfrac x {2^n}}{\dfrac x {2^n}} \prod_{k=0}^n \cos \frac x {2^k} $$

Now let $x=\pi /2$. We get

$$\frac{{\sin \frac{\pi }{2}}}{{\frac{\pi }{2}}} = \frac{{\sin \frac{\pi }{{{2^{n + 1}}}}}}{{\frac{\pi }{{{2^{n + 1}}}}}}\prod\limits_{k = 0}^n {\cos } \frac{\pi }{{{2^{k + 1}}}}$$


$$\frac{2}{\pi } = \frac{{\sin \frac{\pi }{{{2^{n + 1}}}}}}{{\frac{\pi }{{{2^{n + 1}}}}}}\prod\limits_{k = 2}^{n + 1} {\cos } \frac{\pi }{{{2^k}}}$$

  • $\begingroup$ so the limit is \frac{2}{\pi} $\endgroup$ – pourjour Nov 5 '12 at 13:46
  • $\begingroup$ and how did you conclude that \sin x=2^n \sin \frac x {2^n} \prod_{k=1}^n \cos \frac x {2^k}, I know you can use induction to prove but is there any easy way without wasting time looking for this formula $\endgroup$ – pourjour Nov 5 '12 at 13:48
  • $\begingroup$ @pourjour You aren't "wasting time". Induction is the most sensible way I can think about now. Just use the double angle formula to reduce the $n+1$ case to $n$ and you're done. And use dollar signs to get the LaTeX to render. $\endgroup$ – Pedro Tamaroff Nov 5 '12 at 13:51

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