# Using Mean Value Theorem, Prove that ${\tan x\over x}>{x\over\sin x}$

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?

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

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!

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

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 $\displaystyle \frac{\tan x}{x}$ $>1$.

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