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Let x = pi/(2k+1), for k>0. Prove that
cosxcos2xcos3x...coskx = (1/2)^k

I've confirmed this numerically for n from 1 to 30. I'm finding it surprisingly difficult using standard trig formula manipulation. Even for the case k = 2, I needed to actually work out cosx by other methods to get the result.

Please let me know if you have a neat proof.

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Maybe such puzzles are better posted in www.artofproblemsolving.com –  Gerald Edgar Feb 27 '10 at 11:53
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3 Answers

up vote 13 down vote accepted

Let $S(x)=\prod_{j=1}^k \text{sin}(jx)$ and $C(x)=\prod_{j=1}^k \text{cos}(jx)$. Let x = $\frac{\pi}{2k+1}$. Then $S(2x) = S(x)$ (from $\text{sin}(\pi-x)=\text{sin}(x)$), and $S(2x)=2^kS(x)C(x)$ (from $\text{sin}(2x)=2\text{sin}(x)\text{cos}(x)$), from which the result follows.

Steve

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Thanks Steve. Very neat ! –  Cosmonut Feb 27 '10 at 12:03
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Hint: multiply by sin(x)

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The standard way of doing problems like these is to look at the coefficients of the Chebyshev polynomials. The polynomial $T_n$ of degree $n$ such that $T_n(2 \cos \theta) = 2 \cos n \theta$ has leading term $1$, and we want to compute something like the fourth root of the product of the roots of $T_{2k+1}(x)^2 = 4$. Vieta's formulas and some reflection identities should handle it from here.

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