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I am currently working on a challenge problem where I need to show that there is a point $x \in \mathbb{R_+}$ such that $\cos(x) = 0$ using only a few properties of the cosine function. In particular, the only properties of the cosine function that I can use are:

  • $\cos(x)$ is continuous

  • $\cos(x) = Re(\exp(z))$ for $z \in \mathbb{C}$

  • $\cos^2(x) + \sin^2(x)=1$
  • $\displaystyle \cos(x) = \sum_{n=0}^\infty \frac{(-1)^n}{(2n!)}x^{2n}$
  • $\displaystyle \exp(x) = \sum_{n=0}^\infty \frac{x^n}{n!}$

My strategy is to use the intermediate value theorem on the interval $[0,a]$ since it's easy to show that $\cos(0) = 1$. If I could show that there is a point $a \in \mathbb{R}_+$ s.t. $\cos(a) < 0$, then the IVT and the continuity of the cosine function would allow me to conclude that there has to be some $x\in [0,a]$ such that $\cos(x) = 0$.

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See page 182 in Principles of Mathematical Analysis. –  wj32 Nov 1 '12 at 2:37
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Nitpick: your second property should be $\cos x = \Re(\exp(ix))$. –  Javier Badia Nov 1 '12 at 2:56
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2 Answers

up vote 4 down vote accepted

Admittedly not the most elegant of methods. Take the series definition of cosine you have and evaluate $$\cos(2) = \sum_{n=0}^\infty\frac{(-1)^n}{(2n)!}4^n$$ for a few terms.

We know that for an alternating series, we have $|s - s_n| \le a_{n+1}$ where $s$ is the value of the series, $s_n$ the $n$th partial sum and $a_{n+1}$ the absolute value of the $n+1$th term of the series. You can simply evaluate a few terms of the series until you verify that $\cos(2)$ is indeed negative.

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Great minds think alike. –  Ross Millikan Nov 1 '12 at 2:52
    
@RossMillikan Thank you. –  EuYu Nov 1 '12 at 2:56
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I would just use the series for $\cos 2$. You know it is negative, but can't use that. It is an alternating series, so the error is less than the first neglected term and has the same sign. The first three terms suffice.

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