Prove there exists $t>0$ such that $\cos(t) < 0$.

Question

I want to prove that there exists a positive real number $t$ such that $\cos(t)$ is negative.

Here's what I know

$$\cos(x) := \sum_{n=0}^\infty{x^{2n}(-1)^n\over(2n)!}, \;\;(x\in\mathbb R)$$ $${d\over dx}\cos(x) = -\sin(x)$$ $$\cos\left({\pi\over2}\right) = 0, \;\; \cos(0) = 1,$$ $$\sin\left({\pi\over2}\right) = 1, \;\; \sin(0) = 0.$$

I should also specify: The only thing I know about ${\pi\over2}$ is that it is the smallest positive number such that $\cos(\cdot)$ vanishes.

Here's what I tried so far: Since ${d\over dx}\cos(x) = -\sin(x)$ for all $x\in\mathbb R$, we use the fact that $\sin(\pi/2) = 1$ to give us $$\left.{d\over dx}\cos\left(x\right)\right|_{x = {\pi\over2}} = -\sin\left({\pi\over2}\right) = -1.$$ So $\cos(x)$ is decreasing at $x={\pi\over2}$. Using the fact that $\sin(\cdot)$ and $\cos(\cdot)$ are continuous (since differentiable $\implies$ continuous), we know that (and here's where I'm not sure) there exists $\epsilon > 0 $ such that $\cos\left({\pi\over2} + \epsilon\right) < 0$. Call $t = {\pi\over2} + \epsilon > 0$. This completes the proof.

Is there enough justification to make this claim? Since $\sin(\cdot)$ is continuous on $\mathbb R$, for some $\epsilon > 0$ we have $\left.{d\over dx}\cos(x + \epsilon)\right|_{x={\pi\over2}} = -\sin({\pi\over2}+\epsilon) < 0$. Hence $\cos(\cdot)$ is still decreasing at ${\pi\over2} + \epsilon$, and since $\cos({\pi\over2})=0$, it must be that $\cos(t) = \cos({\pi\over2}+\epsilon) < 0 $.

Answer 1

Perhaps a more formal way is to invoke the mean value theorem which says that $$\cos \left( \frac{\pi}{2}+\epsilon \right)-\cos \left( \frac{\pi}{2} \right)=-\sin \left( c \right) \epsilon $$ for some $c \in \left( \frac{\pi}{2},\frac{\pi}{2}+\epsilon \right)$. Using the continuity of $\sin$, you can take $\epsilon$ small enough to guarantee that $\sin(c)$ is sufficiently close to $1$, forcing it to be positive, and therefore finishing the proof.

Answer 2

From the given relations,

$$\cos'\left(\tfrac\pi2\right)=-1=\lim_{h\to0}\frac{\cos\left(\frac\pi2+h\right)-0}{h}.$$

Then by definition of the right-hand limit, taking $\epsilon=1$, there is a $\delta$ such that

$$\forall h:0<h<\delta\implies\left|\frac{\cos\left(\frac\pi2+h\right)}{h}+1\right|<\epsilon=1.$$

This is only possible with negative values.

Answer 3

The terms $a_n$ of the cosine series have alternating signs and (apart from $a_1$) decrease in absolute value when $x=\sqrt{3}$. By the theorem on alternating series it then follows that $$\cos\bigl(\sqrt{3}\bigr)<1-{3\over2}+{9\over24}=-{1\over8}\ .$$

Answer 4

To prove that $\cos\left(\frac{\pi}{2}\right)=0$ from the series definition is just as difficult as proving that $\cos(x)$ is somewhere negative. From the series definition we have that $\cos(0)=1$, $\cos'(0)=0$ (also because $\cos$ is an even function) and that $f(x)=\cos(x)$ is a solution of the differential equation $f''+f=0$. By multiplying both sides by $f'$ and performing termwise integration we get that $f^2+f'^2=1$ (the Pythagorean theorem), in particular $f$ is both bounded and Lipschitz-continuous with a derivative bounded by $1$ in absolute value. By the previous relation $(f,f')$ is directly related with an arc-length parametrization of the unit circle, hence $f$ is periodic and somewhere negative. Also somewhere zero, of course: as a matter of fact, we may define $\pi$ as $$ \pi\stackrel{\text{def}}{=} 2\inf\left\{x\in\mathbb{R}^+:\sum_{n\geq 0}\frac{(-1)^n x^{2n}}{(2n)!}=0\right\} $$ and that is the same as defining $\pi$ as half the length (or the area) of the unit circle.