# upper bound for the product of $\sin (2^k x)$.

Let $n\geq 1$ be an integer and $x$ is a real number. Prove or disprove that :$$\left|\prod_{k=0}^n \sin\left(2^k x\right)\right|\leq\left(\frac{\sqrt{3}}{2}\right)^n.$$

• A hint was attached asking me to find the maximum of $\sin 2t \sin^2 t$ which I found but I could not use. The best thing I could come up with is that the product is less than : $\left(\frac{\sqrt{3}}{2}\right)^n \csc x$. but this does not help. Any hints ? – aziiri Dec 13 '13 at 12:56
• please add that in question in detail.. each and every step... – user87543 Dec 13 '13 at 13:01

## 2 Answers

You should have found that $|\sin 2t\sin^2t|\le \left(\frac{\sqrt3}2\right)^3$. Thus the result follows from $$\left(\prod_{k=0}^n \sin(2^{k}t)\right)^3=\prod_{k=0}^n \sin^3(2^{k}t)=\underbrace{\sin t}_{|\cdot|\le 1}\cdot \prod_{k=0}^{n-1} \underbrace{\sin(2^{k+1}t)\sin^2(2^kt)}_{|\cdot|\le\left(\frac{\sqrt3}2\right)^3}\cdot \underbrace{\sin^2(2^nt)}_{|\cdot|\le 1}$$

This problme is International Mathematics Competition for University Students problem 6,this solution can see: http://www.imc-math.org.uk/imc2001/prob_sol2.pdf