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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.

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    What does "adjacent series" mean?2012-11-05

1 Answers 1

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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

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

or

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

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    @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.2012-11-05