Show that $\cos$ has at least one zeropoint

proof-explanationreal-analysistrigonometry

Suppose $\cos$ has no zeropoints then all values of $\cos$ must be positive due to the fact that $\cos$ is continuous, on $(-\infty,+\infty)$ and $\cos(0)=1$ and $\cos(x)=\cos(-x)$ ,

one could choose any closed intervall $[a,b]$ with $a,b\geq 0$ and assume the opposite that it $\cos(b)$ is negative. The intermediate value Theorem then would say that there must exist a zeropoint.

$(\sin x)'= \cos x$, now because of the Mean Value Theorem and the fact that $\cos>0$. $\sin$ must be strictly rising.

I.e $a_1<x\Longrightarrow \sin a_1< \sin x$

Therefore

$a_1<x\Longrightarrow -\sin a_1>-\sin x$

Because of the fact that $\sin 0 = 0 = -\sin 0 $ and $-\sin x= (\cos x)'$

$(\cos x)'= (-\sin x)\leq (\cos a)' = (-\sin a) < (\cos 0)'=(-\sin 0) = 0$ for $0<a\leq x$

Here is the part now that I don't understand

Why do the Facts so far implicate?

$\cos x\leq \cos a – (x-a)\sin a, \text{(where }\sin a > 0 $ and again for $0<a\leq x\text{)}$

And how does one conclude by that?

$\cos x \rightarrow -\infty$ for $x\rightarrow+\infty$

I understand the last conclusion that if above is true

then it would be a contradiction to $|\cos|\leq 1$ (which we have proved earlier)

Therefore $\cos$ would have at least one zeropoint

Best Answer

You proved that $x\geqslant a\implies\cos'x\leqslant-\sin a$. Now, let $f(x)=\cos(a)-(x-a)\sin(a)$. Then:

  • $f(a)=\cos(a)$;
  • $x\geqslant a\implies f'(x)=-\sin(a)\geqslant\cos'(x)$.

But then $(\forall x\in[a,\infty]):f(x)\geqslant\cos(x)$. So, since $\lim_{x\to\infty}f(x)=-\infty$, $\lim_{x\to\infty}\cos(x)=-\infty$.

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