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Prove that there exists a smallest positive number $p$ such that $\cos(p) = 0 $.

I think I'm supposed to use either Rolle's theorem or the Mean Value Theorem but I'm not sure how, any help would be appreciated. Thanks!

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    Which definition of $\cos x$ are you working with? – 2017-02-21

4 Answers 4

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$\cos$ is a continous function, the set $\{0\}$ is closed, and therefore $\cos^{-1}(\{0\})$ is a closed set.

We conclude that $\cos^{-1}(\{0\})\cap[0,\infty)$ is closed. This set is equal to $\cos^{-1}(\{0\})\cap (0,\infty)$ (because $\cos(0)\neq 0$).

This set is bounded from below and must therefore have a greatest lower bound $\alpha$. Recall that the greatest lower bound of a set is always an adherent point for the set. Since $\cos^{-1}(\{0\})\cap(0,\infty)$ is closed we conclude $\alpha\in \cos^{-1}(\{0\})\cap(0,\infty)$. Therefore the set $\cos^{-1}(\{0\})\cap (0,\infty)$ has a minimum as desired.

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    In general any closed set that is bounded from below has a minimum. – 2017-02-21
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    Maybe “adherent point” rather than “limit point”, but it depends on definitions. The zero set of the cosine consists of isolated points. – 2017-02-21
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I'm not sure the upvoted answer is correct. He/She is implicitly using the assumption that a strictly decreasing sequence of positive real numbers necessarily converges to zero. (i.e. that the strictly decreasing sequence of roots of cos(x) converges to zero). The sequence $$ 1, \frac{1}{1.1}, \frac{1}{1.11}, \frac{1}{1.111}, \frac{1}{1.11111}, \dots $$ is strictly decreasing but is bounded below by $\frac{1}{2}$ so can't converge to zero.

It can be modified as follows. First, since $\cos(0) = 1 > 0$ and $\cos(2) < 0$, the intermediate value theorem tells us that there is a zero of $\cos(x)$ in the interval $(0,2)$. In particular, the set S of positive zeros of the cosine function is nonempty. Also, since we are only considering POSITIVE zeros, this set is bounded below by $0$ and thus has an infimum $t \geq 0$. Choose a sequence $x_n$ of elements of S that converges to $t$. Then by continuity, $$\cos(t) = \cos(\lim x_n) = \lim \cos(x_n) = \lim 0 = 0$$. Thus $\cos(t) = 0$ and since from above $t\geq 0$ and $\cos(0)= 1$ (i.e. not $0$), $\ t > 0$.

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    @Kevin Long Thanks for the information. Am new at this. I added a proof. – 2017-09-12
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There are only finitely many values of $0

One of them must be the smallest

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    How do you prove that there are only a finite number of values? – 2017-02-21
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How about you restrict the domain to I=(0,pi) and show that since cos(x) is monotonic on I, it is one to one, and therefore only one value, p=pi/2, would be the solution on I since cos(a)=cos(b)=0 --> a=b, the solution p in I is unique and since any other positive solution "q" would have to be outside the interval I, pi/2=p