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

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.
