This is one of the problem in my text book where the section which the problem is stated talks about the mean value theorem and Rolle's theorem. By looking at this, I have no idea where to start. Can I have some hints??
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2The derivative of $\cos t$ has absolute value $\le 1$. – André Nicolas Nov 26 '13 at 01:55
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$\cos(x)$ is differentiable in $\mathbb{R}$.
By Mean value theorem $|\cos(x) - \cos(y)| = |sin(\xi)||x - y|$ where $\xi \in (x,y)$ or $(y,x)$.
$|\sin(x)| < 1$ $\forall$ $x \in \mathbb{R}$.
Now get your answer.
Supriyo
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This follows also from the standard $|\sin(x)| \leq |x|$:
$$\left| cos(x) - \cos(y) \right|=2 \left| \sin(\frac{x+y}{2}) \sin(\frac{y-x}{2}) \right| \leq 2 \cdot 1 \cdot \left|\frac{y-x}{2} \right|$$
N. S.
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This is more fundamental and simpler than the other techniques involving differentiation and integration. Differentiation/integration of trig functions are ultimately dependent on $\sin x < x < \tan x$ inequality for $x \in (0, \pi / 2)$. (+1) – Paramanand Singh Nov 26 '13 at 08:04
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On the other hand, $\sin x < x < \tan x$ is a geometric fact, just as the proof for $\sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y)$. Maybe you'd argue you can use complex exponentials, but then you really need complex analysis--i.e. calculus. – abnry Nov 26 '13 at 17:34
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By integration: $$|\cos y - \cos x| = \left|-\int_x^y \sin t dt \right| \leq \int_x^y |\sin t dt| \leq \int_x^y 1 dt = |y-x|$$
abnry
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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ By ${\it Mean Value Theorem}$, $\exists\ \xi \in {\mathbb R}\quad \ni\quad \min\braces{x,y} < \xi < \max\braces{x,y}$ which satisfies $$ {\cos\pars{x} - \cos\pars{y} \over x - y} = -\sin\pars{\xi} \quad\imp\quad \verts{\cos\pars{x} - \cos\pars{y} \over x - y} = \verts{\sin\pars{\xi}} \leq 1 $$
Felix Marin
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