Sine and Cosine are Periodic on Reals/Corollary

Corollary to Sine and Cosine are Periodic on Reals

 * $\cos \left({x + \pi}\right) = - \cos x$
 * $\sin \left({x + \pi}\right) = - \sin x$


 * $\cos x$ is strictly positive on the interval $\left({-\dfrac \pi 2 \,.\,.\, \dfrac \pi 2}\right)$ and strictly negative on the interval $\left({\dfrac \pi 2 \,.\,.\, \dfrac {3 \pi} 2}\right)$


 * $\sin x$ is strictly positive on the interval $\left({0 \,.\,.\, \pi}\right)$ and strictly negative on the interval $\left({\pi \,.\,.\, 2 \pi}\right)$

Proof
From Sine and Cosine are Periodic on Reals:
 * $\sin \left({x + \dfrac \pi 2}\right) = \cos x$
 * $\cos \left({x + \dfrac \pi 2}\right) = -\sin x$

Thus:
 * $\sin \left({x + \pi}\right) = \cos \left({x + \dfrac \pi 2}\right) = -\sin x$
 * $\cos \left({x + \pi}\right) = -\sin \left({x + \dfrac \pi 2}\right) = -\cos x$

It follows directly that:
 * $\forall x \in \left[{-\dfrac \pi 2 \,.\,.\, \dfrac \pi 2}\right]: \cos x \ge 0$

Hence:
 * $\forall x \in \left[{\dfrac \pi 2 \,.\,.\, \dfrac {3 \pi} 2}\right]: \cos x \le 0$

The result for $\sin x$ follows similarly, or we can use:
 * $\sin \left({x + \dfrac \pi 2}\right) = \cos x$