# Thread: proving : tg(x)>x using Lagrange / Cauchy's Lemma Use

1. ## [SOLVED] proving : tg(x)>x using Lagrange / Cauchy's Lemma Use

I need to prove that $\displaystyle tg(x)>x$ when $\displaystyle x \in (0,\frac{\pi}{2})$.

I'll use Cauchy's version for Lagrange's lemma:

If f,g are continuous in [a,b] and derivative in (a,b), then $\displaystyle \frac{f(b)-f(a)}{g(b)-g(a)} = \frac{f'(c)}{g'(c)}$ , (there exists a 'c' : a<c<b for which...)

So I'll take the range [0,x], while $\displaystyle 0<x<\frac{\pi}{2}$, and f(x):=sin(x), g(x):=xcos(x) :

$\displaystyle \frac{tg(x)}{x}=\frac{sinx}{xcosx}=\frac{sin(x)-sin(0)}{xcos(x)-0*cos(0)} = \frac{f'(c)}{g'(c)} = \frac{c}{cos(c)-c*sin(c)}$

Now, how can I prove that the last phrase is larger than 1? (so that $\displaystyle \frac{tg(x)}{x}>1 \Rightarrow tg(x)>x$) ?

Thank you very much!

I need to prove that $\displaystyle tg(x)>x$ when $\displaystyle x \in (0,\frac{\pi}{2})$.

I'll use Cauchy's version for Lagrange's lemma:

If f,g are continuous in [a,b] and derivative in (a,b), then $\displaystyle \frac{f(b)-f(a)}{g(b)-g(a)} = \frac{f'(c)}{g'(c)}$ , (there exists a 'c' : a<c<b for which...)

So I'll take the range [0,x], while $\displaystyle 0<x<\frac{\pi}{2}$, and f(x):=sin(x), g(x):=xcos(x) :

$\displaystyle \frac{tg(x)}{x}=\frac{sinx}{xcosx}=\frac{sin(x)-sin(0)}{xcos(x)-0*cos(0)} = \frac{f'(c)}{g'(c)} = \frac{c}{cos(c)-c*sin(c)}$

Now, how can I prove that the last phrase is larger than 1? (so that $\displaystyle \frac{tg(x)}{x}>1 \Rightarrow tg(x)>x$) ?

Thank you very much!
Don't really think there is any need for Lagrange here...

Let $\displaystyle f(x) = tan(x) -x, ~ x \in (0, \frac{\pi}{2})$, then $\displaystyle f'(x) = \frac{1}{cos^2(x)} - 1$

Now, note that:

1. $\displaystyle f(0) = tan(0) - 0 = 0$

2. Since for any $\displaystyle x \in (0, \frac{\pi}{2}), ~ 0 < cos(x) < 1 \Rightarrow 0 < cos^2(x) < 1 \Rightarrow \frac{1}{cos^2(x)} > 1$, we get that $\displaystyle f'(x) > 0 ~\forall x \in (0, \frac{\pi}{2})$.

These two facts, along with the continuity of $\displaystyle f(x)$, give us that $\displaystyle f(x) > 0 ~ \forall x \in (0, \frac{\pi}{2})$ and therefore $\displaystyle tan(x) > x ~ \forall x \in (0, \frac{\pi}{2})$

3. Originally Posted by Defunkt
Don't really think there is any need for Lagrange here...

Let $\displaystyle f(x) = tan(x) -x, ~ x \in (0, \frac{\pi}{2})$, then $\displaystyle f'(x) = \frac{1}{cos^2(x)} - 1$

Now, note that:

1. $\displaystyle f(0) = tan(0) - 0 = 0$

2. Since for any $\displaystyle x \in (0, \frac{\pi}{2}), ~ 0 < cos(x) < 1 \Rightarrow 0 < cos^2(x) < 1 \Rightarrow \frac{1}{cos^2(x)} > 1$, we get that $\displaystyle f'(x) > 0 ~\forall x \in (0, \frac{\pi}{2})$.

These two facts, along with the continuity of $\displaystyle f(x)$, give us that $\displaystyle f(x) > 0 ~ \forall x \in (0, \frac{\pi}{2})$ and therefore $\displaystyle tan(x) > x ~ \forall x \in (0, \frac{\pi}{2})$
That's a great way to solve it, only I haven't 'officially' learned it, and need to use Lagrange \ Roll \ Cauchy's lemmas for functions. Can you help me?

4. I got it!! I was really close, but I couldn't see it :

I need to prove:

$\displaystyle \frac{c}{cos(c)-csin(c)} > 1$

$\displaystyle \frac{cos(c)-csin(c)}{c} < 1$

$\displaystyle 1-c*tg(c) < 1$

$\displaystyle c*tg(c) > 0$

^ Which is right for every $\displaystyle 0<c<\frac{\pi}{2}$