1. ## Proof

Hi everyone,

I often saw in books :
"Sin(x)<x<Tan(x) thus Cos(x)<Sin(x)/x<1 for (0, π/2)".

So I was wondering how can I show that:

Sin(x)<x<Tan(x) => Cos(x)<Sin(x)/x<1 for (0, π/2)

I tired a few things but it's not very convincing.

2. Originally Posted by sunmalus
I often saw in books :
"Sin(x)<x<Tan(x) thus Cos(x)<Sin(x)/x<1 for (0, π/2)".
So I was wondering how can I show that:
I know that this posted in the Trigonometry Forum but with calculus it is easy.
I assume that this is the part you need to prove, $\displaystyle \sin (x) < x < \tan (x)$.
Observe that $\displaystyle \left( {\forall x \in \left( {0,\frac{\pi }{2}} \right)} \right)\left[ {x > 0,\,\cos (x) > 0,\,\tan (x) > 0} \right]$.
So really all you need prove is $\displaystyle x < \tan (x)$.

Define $\displaystyle f(x)=\tan (x)-x$, then show that $\displaystyle f$ is increasing on set.

3. Hello, sunmalus!

You're looking at the derivative of $\displaystyle \sin x$, aren't you?

This is part of the "Squeeze Theorem" in the derivation.

I was wondering how can I show that:

. . $\displaystyle \sin x\:<\:x\:<\:\tan x \quad\Rightarrow\quad \cos x \:<\:\frac{\sin x}{x} \:<\:1\;\; \text{ for }(0,\:\tfrac{\pi}{2})$

We are given: .$\displaystyle \sin x \:<\: x \:<\:\frac{\sin x}{\cos x}$

Divide by $\displaystyle \sin x\!:\quad 1 \:<\:\frac{x}{\sin x} \:<\:\frac{1}{\cos x}$

Take reciprocals (and reverse the inequalities!)

. . $\displaystyle 1 \:>\:\frac{\sin x}{x} \:>\:\cos x$

Therefore, we have: .$\displaystyle \cos x \:<\:\frac{\sin x}{x} \:<\:1$

4. ## Thanks

Thanks to you both!

But Soroban you really "unstuck" me because I forgot that the reciprocal would change the inequalities, so this is why I wasn't so sure about what I was doing.

Thanks again guys.