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Prove that:

$|\cos(y) - \cos(x)| <|x-y| ,\forall (x,y)\in(\mathbb{R} \times \mathbb{R}) /\triangle(\mathbb{R})$

My attempt

It looks like we need to prove that the cosine function is norm decreasing.So the goal is to show that $$d(f(x),f(y))<d(x,y),\forall x\neq y$$

By the mean value theorem, Let $f(x)=\cos(x)$ then $f'(x)=-\sin(x)$. So $$-\sin(c)= \frac{\cos(y)-\cos(x)}{y-x}$$ $$\Rightarrow (y-x)(- \sin(c))=\cos(y)-\cos(x) $$ I am not sure, but can we conclude anything about cosine being bounded here? Why or why not?

By some trig identities we know that: $$i) \sin(A+B)=\sin(A)cos(B) +\cos(A) \sin (B)$$

$$ii) cos(A+B)= \cos(A) \cos(B) - \sin(A) \sin(B) $$

$$iii) \cos(x)-\cos(y)= -2\sin(\frac{x-y}{2})(\sin\frac{x+y}{2})$$

Plugging $iii$) in $d(f(x)-f(y))<d(x,y)$:

$$|-2\sin(\frac{x-y}{2})(\sin\frac{x+y}{2})|<|x-y|$$ $$= 2|\sin(\frac{x-y}{2})(\sin\frac{x+y}{2})|<|x-y|$$

Don't really know where to go from here

Oh and another thing... why is the domain $\forall (x,y)\in(\mathbb{R} \times \mathbb{R}) /\triangle(\mathbb{R})$, what does this even mean?

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    $\begingroup$ As for your last question, the forward slash seems to mean what is more usually denoted by the backslash $\setminus$, set difference: the expression refers to $(x,y)$ with $x\ne y$, since obviously $\cos(x)-\cos(x)=x-x=0$. $\triangle(\Bbb{R})$ looks like an unusual notation for the diagonal in $\Bbb{R}\times \Bbb{R}$. General advice: do it first for $0\le x < y\le \pi/2$. $\endgroup$ – ForgotALot Apr 29 '17 at 23:34
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The set on which this is true should be written as $$ \mathbb{R} \times \mathbb{R} \setminus \Delta(\mathbb{R}), $$ I believe: this is the set of pairs of real numbers without the diagonal: elements of the form $(x,x)$. $\Delta(\mathbb{R}) = \{(x,y):x=y \} $. This is obviously necessary since both sides are zero if $x=y$, so not unequal.


Pursuing your trigonometric identity idea, you have that $$ \cos{(A+B)} = \cos{A}\cos{B}-\sin{A}\sin{B} \\ \cos{(A-B)} = \cos{A}\cos{B}+\sin{A}\sin{B}, $$ so subtracting gives $$ \cos{(A+B)}-\cos{(A-B)} = - 2\sin{A}\sin{B} . $$ Replacing $A$ and $B$ by $(x\pm y)/2$ then gives $$ \cos{x}-\cos{y} = -2\sin{\left( \frac{x+y}{2} \right)}\sin{\left( \frac{x-y}{2} \right)}. $$ Now, we want to show that the right-hand side is less in absolute value than $\lvert x-y \rvert$. Obviously $\left|\sin{\left( \frac{x+y}{2} \right)}\right|\leq 1$, so it suffices to show that $2\left|\sin{\left( \frac{x-y}{2} \right)}\right|<\lvert x-y \rvert$ for $\lvert x-y \rvert \neq 0$, which can be rewritten as $$ \lvert\sin{u}\rvert<\lvert u \rvert $$ for $u \in \mathbb{R} \setminus \{0\}$. This is obviously true for $\lvert u\rvert>1$ since $\sin{u}$ is bounded between $1$ and $-1$. Moreover, it suffices to check for $u>0$ since $\sin{u}$ is odd. A simple geometrical argument can now be used: draw a circle with centre $O$, and draw two lines through $O$ that contain an angle $u$, meeting the circle at $A$ and $B$ respectively. Then $\triangle AOB$ has area $\frac{1}{2}R^2\sin{u}$, while the sector $OAB$ has area $\frac{1}{2}R^2u$, and is obviously larger than the triangle. The result follows.

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  • $\begingroup$ Thank you, you wrote: $$ \cos{(A+B)} = \cos{A}\cos{B}-\sin{A}\sin{B} \\ \cos{(A+B)} = \cos{A}\cos{B}+\sin{A}\sin{B}, $$ which is a typo? $\endgroup$ – combo student Apr 29 '17 at 23:58
  • $\begingroup$ Of course. Careless copy-and-paste costs lives. Thanks, fixed. $\endgroup$ – Chappers Apr 29 '17 at 23:58
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Hint: Suppose $y>x$. By fundamental theorem of calculus, $\cos(y)-\cos(x)=\int_x^y -\sin(t)\, \textrm{d}t$. It is enough to notice that $ \int_x^y \lvert\sin(t)\rvert \,\textrm{d}t<\int_x^y 1\,\textrm{d}t$.

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