Show that /prod_k=1^n cos (x/2^k) =sin x /(2^n sin(x/2^n)) I have tried with werner on prod_k=1^n cos (x/2^k)sin(x/2^n) and with eulero but it just brings me nowhere. Any suggestion?

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    $\begingroup$ Multiply the left-hand side on the right by $\sin(x/2^n)$. Using the identity $\sin(2t)=2\sin t \cos t$, we get $\left(\prod_1^{n-1}\cos(x/2^k)\right)(1/2)\sin(x/2^{n-1})$. Now use the last sine obtained to "suck up" $\cos(x/2^{n-1})$ and continue. $\endgroup$ Jul 15, 2016 at 7:29
  • $\begingroup$ Thank you Nicolas now is clear $\endgroup$ Jul 15, 2016 at 7:33
  • $\begingroup$ You are welcome. The $\cos(x/2^k)$ terms get chewed up one at a time, leaving a $1/2$ term behind. This continues until the end, when we are left with $\sin x$ multiplied by a bunch of $1/2$. $\endgroup$ Jul 15, 2016 at 7:36


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