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Prove that$$\frac{\sqrt3\cos x-\sin x}{\sin 3x}> \frac{\sqrt3}{3x}-\frac13$$ for small $x$ near $0$.

From Taylor expansion I can see that $$\frac{\cos x}{\sin3x}>\frac13x,\quad \frac{\sin x}{\sin3x}>\frac13,$$ but combining these two does not give actually what I want. I kindly thank you anyone that can provide at least an idea. I know this is true graphically but I want to see rigorously.

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    $\begingroup$ The Maple command series((sqrt(3)*cos(x)-sin(x))/sin(3*x)-(1/3)*sqrt(3)*x+1/3, x, 2); outputs ${\frac {\sqrt {3}}{3}}{x}^{-1}+O \left( x \right) . $ Therefore, the inequality is true for small positive values of $x$ and false for small negative values of $x$. $\endgroup$
    – user64494
    Commented Mar 8, 2018 at 6:45
  • $\begingroup$ Thank you so much for Martin and user64496. I noticed that I did a typo on the right side the inequality the first term should be sqrt(3)/(3x). $\endgroup$
    – user101496
    Commented Mar 8, 2018 at 7:11
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    $\begingroup$ $$\dfrac{\sqrt3\cos x-\sin x}{\sin3x}=\dfrac{2\sin\left(\dfrac\pi3-x\right)}{\sin3\left(\dfrac\pi3-x\right)}$$ $\endgroup$ Commented Mar 8, 2018 at 7:25
  • $\begingroup$ (I second lab bhattacharjee's comment at 2018-03-08 07:25:07Z. The identity is true for every $x\in\mathbb{R}\setminus\{k\frac{\pi}{3}\colon k\in\mathbb{Z}\}$.) $\endgroup$ Commented Mar 8, 2018 at 8:12
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    $\begingroup$ @user101496 I have edited $\frac{\sqrt3}3x$ to $\frac{\sqrt3}{3x}$ - if I understand your comment that you left on the MO version - before migration this is what you intended to ask. Please check whether now the statement of the problem is correct - and if needed, edit the post further to get it to the form you want to sak. $\endgroup$ Commented Mar 8, 2018 at 8:58

2 Answers 2

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Using the identity from @labbhattacharjee, for $0 < x < \dfrac{π}{3}$ there is\begin{align*} 0 < \frac{\sin 3x}{\sqrt{3}\cos x - \sin x} &= \frac{1}{2} \frac{\sin\left( 3\left( \dfrac{π}{3} - x \right) \right)}{\sin\left( \dfrac{π}{3} - x \right)} = \frac{1}{2} \left( 3 - 4\sin^2\left( \dfrac{π}{3} - x \right) \right)\\ &= \frac{1}{2} \left( 1 + 2\cos\left( \dfrac{2π}{3} - 2x \right) \right) = \frac{1}{2} (1 - \cos 2x + \sqrt{3} \sin 2x)\\ &= \sin^2 x + \sqrt{3} \sin x \cos x = \sin x (\sin x + \sqrt{3} \cos x)\\ &< x(x + \sqrt{3}) = x \cdot \frac{3 - x^2}{\sqrt{3} - x} < \frac{3x}{\sqrt{3} - x}, \end{align*} thus$$ \frac{\sqrt{3}\cos x - \sin x}{\sin 3x} > \frac{\sqrt{3} - x}{3x} = \frac{1}{\sqrt{3} x} - \frac{1}{3}. $$

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  • $\begingroup$ I think the x’s in the 4th line should be 2x. $\endgroup$
    – user101496
    Commented Mar 11, 2018 at 8:23
  • $\begingroup$ In the expression of (1-cosx+sqrt(3)sinx)/2 all the x’s should be 2x $\endgroup$
    – user101496
    Commented Mar 11, 2018 at 8:31
  • $\begingroup$ @user101496 Thanks for pointing out! $\endgroup$ Commented Mar 11, 2018 at 8:33
  • $\begingroup$ No problem, the rest follow with the same idea. I really like this proof. Thank you all. $\endgroup$
    – user101496
    Commented Mar 11, 2018 at 8:41
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COMMENT.- $$\frac{\sqrt3\cos x-\sin x}{\sin 3x}> \frac{\sqrt3}{3x}-\frac13\iff\frac{3x}{\sin 3x}\cdot\frac{\sqrt3\cos x-\sin x}{\sqrt3-x}\gt1$$ Both factors in the LHS tend to $1$ and for small $x\gt0$ the first factor is greater than $1$ and the second factor is less than $1$ but it can be proved that for small $x\gt 0$ $$\frac{3x}{\sin 3x}\gt\frac{\sqrt3\cos x-\sin x}{\sqrt3-x}$$ (note that the denominator $\sqrt3-x$ is quite greater than $\sin 3x$ so you look at the numerators)

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  • $\begingroup$ If the first factor is greater than $1$ and the second factor is [less?] than $1$, then $${3x\over\sin3x}\gt1\gt{\sqrt3\cos x-\sin x\over\sqrt3-x}$$ so what's to prove (and how does the hint help)? $\endgroup$ Commented Mar 10, 2018 at 12:45
  • $\begingroup$ Precisely, we would have to prove the last positive difference which could be maybe easier than other way I wanted to indicate a possible way to solve (already $\sin 3x\lt \lt\sqrt3-x$) $\endgroup$
    – Piquito
    Commented Mar 10, 2018 at 14:34
  • $\begingroup$ @Barry Cipra: Thanks for "less" instead of "least" (my English is quite deficient). $\endgroup$
    – Piquito
    Commented Mar 10, 2018 at 14:40

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