Solving x<sin(x)<x w/ Mean Value Theorem

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Homework Statement



-x<sin(x)<x

Homework Equations



show the inequality using the mean value theorem.

The Attempt at a Solution


i try to find c but i keep getting tan(x) as the solution.
 
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i had:

f(x)=sin(x) a=-x b=x

f(x)-f(-x)= f'(c) (x+x)
sin(x) = sin(x)=cos(c) (2x)
2sin(x)=2cos(xc)
tan(x)=c

i don't know if that's right, but i don't get the result.

i would appreciate your help.
 
First, [itex]2x \cos (c) \neq 2 \cos(cx)[/itex] and second [tex]\frac{\cos(cx)}{\cos (x)}\neq c[/tex]!

Try again, but this time use a=0 and b=x.

What do you know about the maximum and minimum values of cosine of any number?
 
ok, now i got:

f(x)=sin(x) a=0 b=x

f(x)-f(0)= f'(c) (x-0)
sin(x) =cos(c) (x)
sin(x)/x=cos(c)

im stucked there...
i don't know what you mean with the the maximum and minimum values of cosine of any number.
 
Well, cosine is a periodic function that is never greater than 1 or less than negative 1...ring a bell?

That means that [itex]-1\leq \cos (c) \leq 1[/itex] and so...