5
$\begingroup$

Prove using Mean Value Theorem,

$${\tan x\over x}>{x\over\sin x} \space\forall \space x \space\epsilon (0, \pi/2) $$

Attempt::

$f(x) = x-\sin x$

$f'(x) = 1-\cos x > 0 $

Hence $x-\sin x>0, {x\over\sin x}>1$

Similarly, I got ${\tan x\over x}>1$

But how do I compare them and get the required inequality?

$\endgroup$
  • $\begingroup$ For $x\in \mathbb R$ or in some interval? (In $\mathbb R$ it does not hold) $\endgroup$ – Jimmy R. Dec 2 '16 at 14:45
  • $\begingroup$ Isn't $x$ restricted to any interval? $\endgroup$ – ajotatxe Dec 2 '16 at 14:45
  • 1
    $\begingroup$ Prove $tanx/x >1$ and $sinx/x<1$ and any interval restriction ? $\endgroup$ – papabiceps Dec 2 '16 at 14:45
  • 1
    $\begingroup$ Interval is between 0<x<pi/2 $\endgroup$ – user3442005 Dec 2 '16 at 15:05
2
$\begingroup$

We need to prove that $f(x)>0$, where $f(x)=\frac{\sin{x}}{\sqrt{\cos{x}}}-x$.

But $f'(x)=\frac{\cos x\sqrt{\cos x}+\frac{\sin^2x}{2\sqrt{\cos x}}}{\cos{x}}-1=\frac{(\sqrt{\cos^3{x}}-1)^2+\cos^2x(1-\cos{x})}{2\sqrt{\cos^3x}}>0$.

Thus, $f(x)>f(0)=0$.

Done!

$\endgroup$
  • $\begingroup$ How is f(x) = sinx/root(cosx) -x ?That isnt the question. $\endgroup$ – user3442005 Dec 2 '16 at 16:58
  • $\begingroup$ @user3442005 We need to prove that $\frac{sin^2x}{\cos{x}}>x^2$, which is $f(x)>0$. $\endgroup$ – Michael Rozenberg Dec 2 '16 at 17:27
  • $\begingroup$ Where did you use Mean Value Theorem? $\endgroup$ – user261263 Dec 3 '16 at 6:05
1
$\begingroup$

For each $x \in (0,2\pi),$ the MVT(mean value theorem) makes sure that there exists $c_x$ with $0 < c_x < x$ such that

$$\frac{\sin x}{x} = \cos c_x < 1.$$

With $f(x) = \sin x$ we have $\sin x = \sin x - \sin 0 = f'(c_x)(x-0)= (\cos c_x)x$

where $0 < c_x < x$.

Now for $\displaystyle \frac{\tan x}{x}>1$

$f(x) = \tan x$ then $f'(x) = \sec^2(x)$. So $\sec^2(x)$ = $\displaystyle \frac{\tan x}{x}$

For each $x \in (0,2\pi)$ $\sec^2(x) >1$. So $\displaystyle \frac{\tan x}{x}$ $>1$.

$\endgroup$
  • $\begingroup$ I had forgotten to mention the limits, it is from (0, pi/2) $\endgroup$ – user3442005 Dec 2 '16 at 15:13
  • $\begingroup$ Then my answer holds true for your given limits. $\endgroup$ – papabiceps Dec 2 '16 at 15:15
  • $\begingroup$ According to the solution, x/sinx>1 and tanx/x>1. How do I compare the two inequalities and get tanx/x > x/sinx? $\endgroup$ – user3442005 Dec 2 '16 at 15:24
  • $\begingroup$ The comparison is with x/sin x, not sin x/x, so this does not solve the problem. $\endgroup$ – marty cohen Dec 2 '16 at 15:30
  • $\begingroup$ $tan$ function is not well defined on $x \in (0,2\pi)$ $\endgroup$ – user261263 Dec 3 '16 at 6:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.