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Thread: Proof

  1. #1
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    Unhappy 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.

    Thanks for your up coming answers
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  2. #2
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    Quote Originally Posted by sunmalus View Post
    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.
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  3. #3
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    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 $

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  4. #4
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    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.
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