Real Cosine Function has Zeroes

Theorem
The cosine function has at least two distinct zeros that are additive inverses of each other.

That is:
 * $\exists \theta \in \R \setminus \set{0}: \map \cos \theta = 0 \wedge \map \cos {-\theta} = 0$

Proof
$\cos x$ is positive everywhere on $\R$.

From Derivative of Cosine Function and Derivative of Sine Function:
 * $\map {D_{xx} } {\cos x} = \map {D_x} {-\sin x} = -\cos x$

Thus $\map {D_{xx} } {\cos x} = -\cos x$ would always be negative.

Thus from Second Derivative of Concave Real Function is Non-Positive, $\cos x$ would be concave everywhere on $\R$.

But from Real Cosine Function is Bounded, $\cos x$ is bounded on $\R$.

By Differentiable Bounded Concave Real Function is Constant, $\cos x$ would then be a constant function.

This contradicts the fact that $\cos x$ is not a constant function.

Thus, by Proof by Contradiction, $\cos x$ cannot be positive everywhere on $\R$.

Therefore $\exists \phi \in \R: \cos \phi < 0$.

From Cosine of Zero is One we have that $\cos 0 = 1 > 0$.

Since the Cosine Function is Continuous, by the Intermediate Value Theorem, $\exists \theta \in \R: \cos \theta = 0$.

Because Cosine of Zero is One, $\theta \neq 0$.

From Cosine Function is Even, it follows that $\map \cos {-\theta} = 0$.