A trigonometry problem - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T00:42:16Z http://mathoverflow.net/feeds/question/16583 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/16583/a-trigonometry-problem A trigonometry problem Cosmonut 2010-02-27T08:48:16Z 2010-02-27T11:50:47Z <p>Let x = pi/(2k+1), for k>0. Prove that<br> cosxcos2xcos3x...coskx = (1/2)^k</p> <p>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.</p> <p>Please let me know if you have a neat proof.</p> http://mathoverflow.net/questions/16583/a-trigonometry-problem/16589#16589 Answer by ambrosiac for A trigonometry problem ambrosiac 2010-02-27T11:30:58Z 2010-02-27T11:30:58Z <p>Hint: multiply by sin(x)</p> http://mathoverflow.net/questions/16583/a-trigonometry-problem/16591#16591 Answer by Steve D for A trigonometry problem Steve D 2010-02-27T11:39:23Z 2010-02-27T11:39:23Z <p>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.</p> <p>Steve</p> http://mathoverflow.net/questions/16583/a-trigonometry-problem/16592#16592 Answer by Qiaochu Yuan for A trigonometry problem Qiaochu Yuan 2010-02-27T11:44:40Z 2010-02-27T11:50:47Z <p>The standard way of doing problems like these is to look at the coefficients of the <a href="http://en.wikipedia.org/wiki/Chebyshev_polynomials" rel="nofollow">Chebyshev polynomials</a>. 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.</p>