# mean value theorem: $| \cos x-1 | \leq | x |$

I've been studying mean value theorem, and I was told that I should use MTV for solving problems with $\leq$. So I've been trying to show that $| \cos x - 1 | \leq | x |$ for all x values, using MTV, but I just don't get how MTV can be used...
I've been stuck on this for days now and I would be really grateful for any help.

For $x\ne0$ you should know that MVT says $$\frac{\cos x-\cos0}{x-0}=\cos'\xi$$ for some $\xi\in(0,x)$ or $(x,0)$ depending on the sign of $x$.

Therefore $|\cos x-1|=|\cos x-\cos 0|=|x-0||\cos'\xi|=|x||\!\!-\!\sin\xi|\le|x|\cdot1=|x|.$

Verification for $x=0$ is trivial.

• Thank you!! I think I'm finally understanding this... – user125342 Mar 31 '14 at 16:09

$$|\cos x-1|=|\cos x-\cos0|=|\cos'(\xi)(x-0)|\le\ ?$$ $\xi$ is an intermediate point between $0$ and $x$.

Put $f(x):= \cos(x)$. Then $|f'(x)| = |-\sin(x)| \leq 1 \quad \forall x \in \mathbb{R}$.

For any $x > 0$ there exists (by the MVT) a $\theta \in (0,x)$ such that

$$f(x) - f(0) = f'(\theta)(x-0)$$

or in other words

$$\cos(x) - 1 = f'(\theta)x.$$

Taking the modulus gives

$$|\cos(x) - 1| = |f'(\theta)| \cdot |x| \leq 1 \cdot |x| = |x|.$$

You can deal with the case $x<0$ similarly. The case $x=0$ is trivial.