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If $x \in (0,\frac \pi6)$, then using calculus prove that $x\csc x<\frac \pi3$

My attempt:

let $f(x)=\csc x$$$\implies f'(x)=-\frac{\cos{x}}{\sin^2 x}$$ which is less than $0$ for all $x\in (0,\pi/6)$

so, $f'(x)<0$$\implies f(x)$ is decreasing function.

(as we know decreasing function reverses inequality)

$$\csc(0)>\csc(x)>2$$ (notice that inequality is reversed)and since $x$ is positive $$2x<x\csc x<\infty$$but this violates the question and i am definitely wrong somewhere but i don't know where!

Any Help will be appreciated :-D

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  • $\begingroup$ @DanielFischer yes! but i am curious what's wrong in this method. $\endgroup$
    – user5954246
    May 12, 2016 at 20:21
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    $\begingroup$ It doesn't violate the asserted inequality, it's just weaker. Much weaker. Look whether differentiating $\frac{x}{\sin x}$ helps. $\endgroup$
    – Daniel Fischer
    May 12, 2016 at 20:21
  • $\begingroup$ @DanielFischer what's that "weaker"? $\endgroup$
    – user5954246
    May 12, 2016 at 20:24
  • $\begingroup$ $a < 3$ is a weaker inequality than $a < 2$. If you want to prove the latter, but your proof only proves the former, you have proved a too weak inequality. That doesn't imply that the stronger inequality doesn't hold, though. $\endgroup$
    – Daniel Fischer
    May 12, 2016 at 20:28
  • $\begingroup$ @DanielFischer but my equality is not weaker it completely violates, question says to prove $x\csc x<\pi/3$ but what i proved is $x\csc x$ is greater than $2$ which is false! $\endgroup$
    – user5954246
    May 12, 2016 at 20:30

1 Answer 1

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Let $$ f(x) = x \csc (x) = \frac{-2x\sin x}{\cos (2x) - 1},\ x \in ]0, \pi/6 [ $$ $f$ is continous everywhere on the interval, so there exists a max/min.

$$ f'(x) = \frac{-2x \sin (2x) \sin (x)}{(\cos(2x) -1 )^2} - \frac{2 \sin (x)}{\cos (2x) - 1} = -\frac{-2\sin (x)}{\cos (2x) - 1} \left( \frac{x\sin (2x)}{\cos (2x) - 1} + 1 \right) > 0 $$ so $f$ is increasing, max at $x \to \pi/6$ calculating $f(\pi/6) = \pi/3 \Rightarrow$ inequality holds.

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